Cap19

Chapter 19 The Second Law of ThermodynamicsConceptual Problems 1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the hightemperature reservoir) is the ratio of Qc to Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to find Qc/ Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Solving for Qc/ Qh yields: ε= Q W Qh − Qc = = 1− c Qh Qh Qh Qc = 1− ε Qh Substitute for ε to obtain: Qc = 1 − 0.25 = 0.75 Qh and (c ) is correct. 2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the efficiency of the heat engine in terms of Qc and W. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: ε= W W = = Qh W + Qc 1 Q 1+ c W 1779 1780 Chapter 19 Substitute for Qc and W to obtain: 1 = 0.2 400 kJ 1+ 100 kJ and (a ) is correct. ε= 3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Substitute for Qc and Qh to obtain: ε= Q W Qh − Qc = = 1− c Qh Qh Qh 480 kJ = 0.2 600 kJ and (a ) is correct. ε = 1− 4 • Explain what distinguishes a refrigerator from a ″heat pump.″ Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pumps is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter. An air conditioner’s COP is mathematically identical to that Q of a refrigerator, that is, COPAC = COPref = c . However a heat pump’s COP is W Q defined differently, as COPhp = h . Explain clearly why the two COPs are W defined differently. Hint: Think of the end use of the three different devices. 5 • [SSM] Determine the Concept The COP is defined so as to be a measure of the The Second Law of Thermodynamics 1781 effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. • Explain why you cannot cool your kitchen by leaving your refrigerator 6 door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?) Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room, so the refrigerator actually heats the room in which it is located. The heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside. 7 • Why do steam-power-plant designers try to increase the temperature of the steam as much as possible? Determine the Concept Increasing the temperature of the steam increases its energy content. In addition, it increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 8 • To increase the efficiency of a Carnot engine, you should (a) decrease the temperature of the hot reservoir, (b) increase the temperature of the cold reservoir, (c) increase the temperature of the hot reservoir, (d) change the ratio of maximum volume to minimum volume. Determine the Concept Because the efficiency of a Carnot cycle engine is given T by ε C = 1 − c , you should increase the temperature of the hot reservoir. (c ) is Th correct. 9 •• [SSM] Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it requires less work to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. A Carnot engine operates between a cold temperature reservoir of 10 •• 1782 Chapter 19 27°C and a high temperature reservoir of 127°C. Its efficiency is (a) 21%, (b) 25%, (c) 75%, (d) 79%. Determine the Concept The efficiency of a Carnot cycle engine is given by ε C = 1 − Tc Th where Tc and Th (in kelvins) are the temperatures of the cold and hot reservoirs, respectively. Substituting numerical values for Tc and Th yields: εC = 1− 300 K = 0.25 400 K (b ) is correct. 11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is (a) 0.33, (b) 1.3, (c) 3.0 (d) 4.7. Determine the Concept The coefficient of performance of a Carnot cycle engine Q run in reverse as refrigerator is given by COPref = c . We can use the relationship W between W, Qc, and Qh to eliminate W from this expression and then use the Q T relationship, applicable only to a device operating in a Carnot cycle, c = c to Qh Th express the refrigerator’s COP in terms of Tc and Th. The coefficient of performance of a refrigerator is given by: Qc W or, because W = Qh − Qc , COPref = COPref = Qc Qh − Qc Dividing the numerator and denominator of this fraction by Qc yields: For a device operating in a Carnot cycle: Substitute in the expression for COPref to obtain: COPref = 1 Qh −1 Qc Qc Tc = Qh Th COPref, C = 1 Th −1 Tc Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. (b) remains constant. (b) ΔSA > ΔSB. its entropy decreases (the liquid state is a more ordered state than is the vapor state). A = ΔEint. Identify this cycle and sketch it on a PV diagram. . (d) may decrease or remain unchanged. B .The Second Law of Thermodynamics 1783 Substitute numerical values and evaluate COPref. (a) ΔEint A > ΔEint B. the entropy of the water (a) increases. It is not dependent on the process by which the change occurs. Determine the Concept The two paths are shown on the PV diagram to the right. C = 12 •• On a humid day. During condensation. water vapor condenses on a cold surface. like Eint. Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST 14 •• diagram. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem. COPref. Pf Pi P B Ti B f i A Vi Vf A Tf V (a) Because Eint is a state function and the initial and final states are the same for the two paths.0 400 K −1 300 K (c ) is correct. ΔEint. Tf. (c) ΔSA < ΔSB. Thus ΔS A = ΔS B . 13 •• An ideal gas is taken reversibly from an initial state Pi. Explain your answer. (b) and (c) S. (d) (d ) is correct. (c) is correct. Vf. For these two paths. (c) decreases. (d) None of the above. Vi. Ti to the final state Pf. is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. Determine the Concept When water vapor condenses. C: 1 = 3. Therefore. in Figure 19-3. ΔS = ΔEint = Qin = Q = C V ΔT = C V (T − Tb ) C V (T − Tb ) ⎛ T ⎞ = C V ⎜1 − b ⎟ T ⎝ T ⎠ ΔS = . Refer to Figure 19-3. To determine how S depends on T along b→c and d→a. Determine the Concept Note that A→B is an adiabatic expansion. heat is added to the system and both S and T increase. Q is positive. Sketch an ST diagram of the Otto cycle. respectively: Because Won = 0 for this constantvolume process: Substituting for Q yields: Q T where. a. The cycle is that of the Otto engine (see Figure 19-3). Identify the type of engine represented by this diagram. B→C is a constant-volume process in which the entropy decreases. d. from c→d S is constant while T decreases. and from d to a both S and T decrease. P B C A D V 15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV diagram. So.1784 Chapter 19 Determine the Concept The processes A→B and C→D are adiabatic and the processes B→C and D→A are isothermal. because heat is entering the system. respectively. from b to c. the cycle is the Carnot cycle shown in the adjacent PV diagram.) Determine the Concept The Otto cycle consists of four quasi-static steps. (The Otto cycle is discussed 16 •• in Section 19-1. and D in Figure 19-13 correspond to points c. b→c is a constant volume heating. C. consider the entropy change of the gas from point b to an arbitrary point on the path b→c where the entropy and temperature of the gas are S and T. B. from a to b. The points A. c→d is an adiabatic expansion and d→a is a constant-volume cooling. C→D is an adiabatic compression and D→A is a constant-volume process that returns the gas to its original state. and b. S is constant and T increases. There a→b is an adiabatic compression. process 1→2 is an isothermal expansion. S d c a b T 17 •• [SSM] Sketch an SV diagram of the Carnot cycle for an ideal gas. along path b→c.): These results tell us that. Process 3→4 is an isothermal compression in which S decreases and V also decreases.The Second Law of Thermodynamics 1785 On path b→c the entropy is given by: ⎛ T ⎞ S = S b + ΔS = S b + C V ⎜1 − b ⎟ ⎝ T ⎠ T dS = C V b2 dT T 2 T d S = −2C V b3 2 dT T The first and second derivatives. In this process heat is added to the system and the entropy and volume increase. that is. Calculate these derivatives assuming CV is constant. Determine the Concept Referring to Figure 19-8. An ST diagram for the Otto cycle is shown to the right. dS dT and d 2 S dT 2 . . so S is constant as V increases. process 4→1 is adiabatic. and S is constant while V decreases. the slope of the path is positive and the slope decreases as T increases. along path d→a. the slope of the path is positive and the slope decreases as T increases. isentropic. Finally. The concavity of the path is negative for all T. Following the same procedure on path d→a gives: ⎛ T ⎞ S = S d + C V ⎜1 − d ⎟ ⎝ T ⎠ T dS = C V d2 dT T 2 T d S = −2C V d3 2 dT T These results tell us that. (For an ideal gas CV is a positive constant. Process 2→3 is adiabatic. The concavity of the path is negative for all T. give the slope and concavity of the path. ) . S V An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right. The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by: For an isothermal expansion. S 2 3 1 4 V 18 •• Sketch an SV diagram of the Otto cycle. we have: The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. are given by: Substituting for Q yields: ΔS = Q T1 ⎛V ⎞ Q = W = nRT1 ln⎜ ⎜V ⎟ ⎟ ⎝ 1⎠ ⎛V ⎞ ΔS = nR ln⎜ ⎟ ⎜V ⎟ ⎝ 1⎠ ⎛V ⎞ S = S1 + nR ln⎜ ⎜V ⎟ ⎟ ⎝ 1⎠ Since S = S1 + ΔS . (The Otto cycle is discussed in Section 19-1.1786 Chapter 19 During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. This graph establishes the curvature of the 1→2 and 3→4 paths for the SV graph. and thus the heat added to the gas. the work done by the gas. The cycle is shown in the adjacent PV diagram. ″But that can’t happen. While it is true that systems tend to degenerate to greater levels of disorder. it is an adiabatic process in which the pressure increases. Make a sketch of this cycle on a PV diagram. P C B D A V 20 •• One afternoon. or is cleaning the room really impossible? Determine the Concept The son is out of line. A sketch of the SV diagram for the Otto cycle follows: S c d b a V 19 •• Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP diagram. Q > 0 and. it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. it is not true that order cannot be brought forth from disorder. while ΔV = 0 along this path. Mom. process d→a is a constant-volume process. he’s also wrong. heat is exhausted from the system and the volume decreases. Process C→D is an adiabatic compression. here. and your friend replies. Finally. Process B→C is one in which P is constant and S decreases. this time with heat leaving the system and ΔS < 0. ″Well. That’s all.″ Critique your friend’s response. Is his mother correct to ground him for not cleaning his room. Process c→d also takes place adiabatically and so. therefore ΔS > 0. both Q = 0 and ΔS = 0 along this path. Process b→c takes place at constant volume. however. It would violate the second law of thermodynamics. but besides that. again. the mother of one of your friends walks into his room and finds a mess. She asks your friend how the room came to be in such a state. Process D→A returns the system to its original state at constant pressure. Qin. Determine the Concept Process A→B is at constant entropy.″ Her reply is a sharp ″Nevertheless. is positive and so.The Second Law of Thermodynamics 1787 Determine the Concept The Otto cycle is shown in Figure 19-3. that is. you’d better clean your room!″ Your friend retorts. What is . Process a→b takes place adiabatically and so both Q = 0 and ΔS = 0 along this path. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease. The ratio of the coefficients of performance in the basement and kitchen is given by: Qc.basement Qc.kit and simplify to obtain: .kit − Qc.basement W = c.kit Wc.kit −1 Qc.kit Qh. If we further assume that the freezer operates in a Carnot cycle.kit 1 COPbasement COPkit Qh. Estimation and Approximation Estimate the change in COP of your electric food freezer when it is 21 • removed from your kitchen to its new location in your basement. your friend – on the system in order to reduce the level of chaos and bring about order.kit Qc.basement = 1 Qh. which is 8°C cooler than your kitchen. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen.basement Qc.basement −1 Qc.kit Qh.basement and Qc.1788 Chapter 19 required is an agent doing work – for example. and freezer.kit −1 Qc.basement COPbasement COPkit Because W = Qh − Qc for a heat engine or refrigerator: COPbasement COPkit Divide the numerators and denominators by Qc.basement Q − Qc. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task. then we can use the proportion Qh Qc = Th Tc to express the ratio of the coefficients of performance in terms of the temperatures in the kitchen. basement.kit = Qh.basement = h.basement −1 Qc. If the original volume of the air in your bedroom is V1. then Qh Th = and our expression for the Qc Tc ratio of the COPs becomes: Assuming that the temperature in your kitchen is 20°C and that the temperature of the interior of your freezer is −5°C.47 = 285 K COPkit −1 268 K or an increase of 47% in the performance of the freezer! 22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room. ⎛1⎞ p=⎜ ⎟ ⎝ 10 ⎠ N= PV kT N N (1) Use the ideal-gas law to express N: Substitute numerical values and evaluate N: ( 101. substitute numerical values and evaluate the ratio of the coefficients of performance: Th.252 × 10 27 molecules .381×10 −23 J/K )(293 K ) = 1.basement COPkit −1 Tc. confined to your closet whose volume is V2 is given by: N ⎛ V2 ⎞ p=⎜ ⎟ ⎜V ⎟ ⎝ 1⎠ 1 or. normally in your bedroom.basement 293 K −1 COPbasement 268 K = 1. respectively. because V2 = 10 V1 . We can use the ideal-gas law to find the number of molecules N.kit −1 Tc. the probability p of finding the N molecules. We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20°C.325 kPa )(50 m 3 ) N= (1. Picture the Problem The probability that all the molecules in your bedroom are ⎛ V2 ⎞ located in the (open) closet is given by p = ⎜ ⎜V ⎟ ⎟ where N is the number of air ⎝ 1⎠ molecules in your bedroom and V1 and V2 are the volumes of your bedroom and closet.The Second Law of Thermodynamics 1789 If we assume that the freezer unit operates in a Carnot cycle.kit COPbasement = Th. 1790 Chapter 19 Substitute for N in equation (1) and evaluate p: ⎛1⎞ p=⎜ ⎟ ⎝ 10 ⎠ 1. ) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. . your physics professor comes into your store to buy a new refrigerator.0)1. Wanting to buy the most efficient refrigerator possible. One day. Assume the engine operates according to the Otto cycle and assume γ = 1.4−1 24 •• You are working as an appliance salesperson during the summer.4 for diatomic gases) and evaluate εC: 1 ≈ 56% (8. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th: Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasistatic adiabatic compression from Vc to Vh: Substitute for Tc to obtain: Th εC = 1 − Tc Th T V γ −1 ⎛ Vh ⎞ ⇒ c = hγ −1 = ⎜ ⎟ ⎜V ⎟ Th Vc ⎝ c⎠ γ −1 TcVc γ −1 = ThVh γ −1 ⎛ Vh ⎞ εC = 1 − ⎜ ⎟ ⎜V ⎟ ⎝ c⎠ γ −1 Express the compression ratio r: r= Vc Vh 1 Substituting for r yields: εC = 1 − ε C = 1− r γ −1 Substitute numerical values for r and γ (1. she asks you about the efficiencies of the available models.0:1.0.252×10 27 = 10 −1. To make the sale. (The Otto cycle is discussed in Section 19-1. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power.4. She decides to return the next day to buy the most efficient refrigerator. you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator.252×10 27 ≈ 10 −10 23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.252×10 27 = 27 1 101. (a) Using its definition. for the sake of finding the upper limit on the COP. then the refrigerator must be able to maintain a temperature difference of 20 K.65)(600 J/s ) = 8.The Second Law of Thermodynamics 1791 Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0°C (273 K) and the room temperature to be 20°C (293 K). express the COP of a household refrigerator: Apply conservation of energy to the refrigerator to obtain: Substitute for W and simplify to obtain: COP = Qc W (1) W + Qc = Qh ⇒ W = Qh − Qc Qc 1 = Qh − Qc Qh −1 Qc COP = Assume.65 ≈ 14 293 K −1 273 K (2) Substitute numerical values and evaluate COPmax: COPmax = (b) Solve equation (1) for Qc: Differentiate equation (2) with respect to time to obtain: Substitute numerical values and dQc evaluate : dt Qc = W (COP ) dQc dW = (COP ) dt dt dQc = (13. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator.2 kW dt . In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment. that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Qh and Qc: Substitute for Qh Th = Qc Tc Qh to obtain: Qc COPmax = 1 Th −1 Tc 1 = 13. how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe.1792 Chapter 19 25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K. and (d) 1.75 × 1017 W dS Earth P = dt TEarth (b) Express the rate at which Earth’s entropy SEarth changes due to the flow of solar radiation: Substitute numerical values and dS Earth evaluate : dt dS Earth 1. which is about 1010 years.37 kW/m2. the average temperature of the surface of Earth is about 290 K.746 × 1017 W = 1.0 mole. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation. (c) 1000. If a vacuum chamber has the same volume as the box. (a) Estimate the total power of the sunlight hitting Earth. Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting Earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun. We don’t care which side the molecules are on as long as they all are on one side. (a) Using its definition. Picture the Problem If you had one molecule in a box. (e) The best vacuums that have been created to date have pressures of about 10–12 torr.0-L box contains N molecules of an ideal gas. express the intensity of the Sun’s radiation on Earth in terms of the power P delivered to Earth and Earth’s cross sectional area A: Solve for P and substitute for A to obtain: Substitute numerical values and evaluate P: I= P A P = IA = IπR 2 where R is the radius of Earth.02 ×1014 J/K ⋅ s 26 •• A 1.746 ×1017 W = dt 290 K = 6. and the positions of the molecules are observed 100 times per second.37 kW/m 2 6. (b) 100.37 × 10 6 m ( )( ) 2 = 1. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to (a) 10. P = π 1. it would have a 50% chance of being on one side or the other. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1. so with one molecule you have a 100% . one on one side and one on the other.022×10 t= 2(100 s −1 ) 23 (6. In (e) we can apply the ideal gas law to find the number or the other is t = 2(100 ) of molecules in 1.0 mol =6. or a 50% chance of them both being on either side. Extending this logic. (a) Evaluate t for N = 10 molecules: t= 210 = 5.156 × 10 7 s = 0. which means that.The Second Law of Thermodynamics 1793 chance of it being on one side or the other.0 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K. the probability of N molecules all being on one side of the box is P = 2/2N.022 ×10 )ln 2 = x ln10 ⇒ x ≈ 10 23 23 . the time it would take them to cover all the combinations and all get on one side 2N . there are four possible combinations (both on one side.5 × 10 299 s × ≈ 2 × 10 291 y (d) Evaluate t for N = 1.12 s ≈ 5 s 2 100 s −1 ( ) (b) Evaluate t for N = 100 molecules: t= 2100 2 100 s −1 ( ) 1y 3. so there is a 25% (1 in 4) chance of them both being on a particular side.156 × 10 7 s = 6. both on the other. and the reverse). if the molecules shuffle 100 times a second.022×10 let 23 10 x = 26. With two molecules.34 × 10 27 s × ≈ 2 × 10 20 y (c) Evaluate t for N = 1000 molecules: t= 21000 2 100 s −1 ( ) To evaluate 21000 let 10 x = 21000 and take the logarithm of both sides of the equation to obtain: (1000)ln 2 = x ln10 ⇒ x = 301 Substitute to obtain: t= 10 301 2 100 s −1 ( ) 1y 3.022 ×1023 molecules: To evaluate 26.022×10 and take the logarithm of both sides of the equation to obtain: 23 26. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? .0% efficiency does 0.22×10 and take the logarithm of both sides of the equation to obtain: Substituting for x yields: 7 N= PV kT −12 (10 N= torr )(133.32 Pa/torr )(1.381×10−23 J/K )(300 K ) = 3.22×10 let 7 10 x = 23.156 ×10 s ⎠ 23 ( ) ≈ 1010 y 23 (e) Solve the ideal gas law for the number of molecules N in the gas: Assuming the gas to be at room temperature (300 K).22 × 10 )ln 2 = x ln10 ⇒ x ≈ 10 7 7 1010 1y × t= −1 2 100 s 3.100 kJ of work during each cycle.0 L ) (1.1794 Chapter 19 Substituting for x yields: 1010 1y ⎛ ⎞ t≈ 7 ⎟ −1 ⎜ 2 100 s ⎝ 3.22×10 t= 2(100 s −1 ) 7 (3. substitute numerical values and evaluate N: Evaluate t for N = 3.22 × 107 molecules 2 3.156 × 10 7 s 7 ( ) ≈ 1010 y 7 Express the ratio of this waiting time to the lifetime of the universe tuniverse: t tuniverse or 7 1010 y = 10 ≈ 1010 10 y 7 t ≈ 1010 tuniverse 7 Heat Engines and Refrigerators 27 • [SSM] A heat engine with 20.22×107 molecules: To evaluate 23. (a) What is its efficiency? (b) If each cycle takes 0.200 ε Qc = Qh − W = 500 J − 100 J = 400 J 28 • A heat engine absorbs 0. (a) The efficiency of the heat engine is given by: (b) Apply conservation of energy to the engine to obtain: Substitute numerical values and evaluate Qc: ε= W 120 J = = 30% Qh 400 J Qh = W + Qc ⇒ Qc = Qh − W Qc = 400 J − 120 J = 280 J 29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. find the power output of this engine.400 kJ of heat from the hot reservoir and does 0. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output. we can express the efficiency of the engine in terms of the heat Qc released to the cold reservoir during each cycle. Qh = W + Qc .120 kJ of work during each cycle. (a) Qh absorbed from the hot reservoir during each cycle is given by: (b) Use Qh = W + Qc to obtain: Qh = W = 100 J = 500 J 0.The Second Law of Thermodynamics 1795 Picture the Problem (a) The efficiency of the engine is defined to be ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) Because. from conservation of energy. (b) We can apply conservation of energy to the engine to obtain Qh = W + Qc and solve this equation for the heat Qc released to the cold reservoir during each cycle. (a) The efficiency of the heat engine is given by: ε= Q W Qh − Qc = = 1− c Qh Qh Qh .50 s. 0 kJ of heat from a cold reservoir and releases 8.500 s ⎟ ⎟ = 80 W ⎝ ⎠ 30 • A refrigerator absorbs 5. what is its efficiency? Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.0 kJ = 1.0 kJ = 38% 8.0 kJ ε= W Qh Qh − Qc Q = 1− c Qh Qh ε= ε = 1− 5.0 kJ to a hot reservoir.7 8. If it is run backward as a heat engine between the same two reservoirs.0 kJ − 5.40 )⎜ ⎜ 0. (a) The COP of a refrigerator is defined to be: Apply conservation of energy to relate the work done per cycle to Qh and Qc: Substitute for W to obtain: COP = Qc W W = Qh − Qc COP = Qc Qh − Qc Substitute numerical values and evaluate COP: (b) The efficiency of a heat pump is defined to be: Apply conservation of energy to the heat pump to obtain: Substitute numerical values and evaluate ε : COP = 5.0 kJ . (a) Find the coefficient of performance of the refrigerator. (b) The refrigerator is reversible.1796 Chapter 19 Substitute numerical values and evaluate ε: (b) The power output P of this engine is the rate at which it does work: Substitute numerical values and evaluate P: ε = 1− 60 J = 40% 100 J P= dQh dW d = ε Qh = ε dt dt dt ⎛ 100 J ⎞ P = (0. It is then compressed at constant pressure to its original state. find the work done by the gas. Hence: T2 = 2T1 = 600 K T3 = 2T2 = 1200 K . (a) Show this cycle on a PV diagram.00 atm and V1 = 24.6 L. (a) The cycle is shown to the right: Apply the ideal-gas law to state 1 to find T1: T1 = P1V1 = nR (1. we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle.2 L. Finally.00 mol of a monatomic ideal gas. 3. We can use the 1st law of thermodynamics to find the change in internal energy for each step of the cycle. The cycle begins at P1 = 1. All the steps are quasi-static and reversible. Picture the Problem To find the heat added during each step we need to find the temperatures in states 1. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q = CV ΔT and Q = CP ΔT . The gas is heated at constant volume to P2 = 2. 2.00 atm )(24.00 atm.00 mol)⎛ ⎜ 8. It then expands at constant pressure until its volume is 49. and 4. Hence: The volume doubles while the pressure remains constant between states 2 and 3. (b) Find the efficiency of the cycle. For each step of the cycle.6 L ) − 2 L ⋅ atm ⎞ (1. and the change in the internal energy of the gas.The Second Law of Thermodynamics 1797 31 •• [SSM] The working substance of an engine is 1.00 atm. The gas is then cooled at constant volume until its pressure is again 1.206 ×10 ⎟ mol ⋅ K ⎠ ⎝ = 300 K The pressure doubles while the volume remains constant between states 1 and 2. the heat absorbed by the gas. 34 = −7.6 L )⎜ ⎟ = − 4.48 kJ 2 2⎜ mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain: ΔEint. 23 = 12.99 kJ ⎝ L ⋅ atm ⎠ J ⎞ ⎛ Q23 = CP ΔT23 = 5 RΔT23 = 5 8.48 kJ + 0 = − 7.74 kJ 2 2⎜ mol ⋅ K ⎠ ⎝ The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics: Because W12 = 0 : For path 2→3: ΔEint = Qin + Won ΔEint.5 kJ and J ⎞ ⎛ Q34 = ΔEint.314 ⎟ (600 K − 1200 K ) = − 7.74 kJ ⎛ 101. Hence: For path 1→2: W12 = PΔV12 = 0 T4 = 1 T = 600 K 2 3 and J ⎞ ⎛ Q12 = CV ΔT12 = 3 RΔT12 = 3 8.314 ⎟ (1200 K − 600 K ) = 12.12 = Q12 = 3.314 ⎟ (600 K − 300 K ) = 3.99 kJ = 7.00 atm )(49.48 kJ .1798 Chapter 19 The pressure is halved while the volume remains constant between states 3 and 4.325 J ⎞ Won = −W23 = − PΔV23 = −(2. 34 = CV ΔT34 = 3 RΔT34 = 3 8.5 kJ 2⎜ 2 mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain: For path 3→4: W34 = PΔV34 = 0 ΔEint.5 kJ − 4.2 L − 24. 99 kJ − 2.2 L )⎜ ⎟ = 2. .24 3.5 −7.The Second Law of Thermodynamics 1799 For path 4→1: ⎛ 101.00 atm and a volume of 20.74 kJ + 12.0 L.74 12. Note also that. kJ 1→2 2→3 3→4 4→1 0 −4. kJ Qin . kJ ΔEint (= Qin + Won ) .49 kJ ⎝ L ⋅ atm ⎠ and J ⎞ ⎛ Q41 = CP ΔT41 = 5 RΔT41 = 5 8. 41 = −6. 32 •• The working substance of an engine is 1.5 kJ Remarks: Note that the work done per cycle is the area bounded by the rectangular path.00 mol of a diatomic ideal gas.0 L to a pressure of 1. the sum of the changes in the internal energy for the cycle is zero.24 kJ + 2.24 kJ 2⎜ 2 mol ⋅ K ⎠ ⎝ Apply ΔEint = Qin + Won to obtain: ΔEint.99 0 2.48 −6. the results of the preceding calculations are summarized in the following table: Process Won .314 ⎟ (300 K − 600 K ) = − 6. (2) a compression at constant pressure to its original volume of 10.49 3.48 −3.49 kJ ≈ 15% 3.325 J ⎞ Won = −W41 = − PΔV41 = −(1.0 L.6 L − 49.74 7.00 atm )(24.5 −7.75 kJ For easy reference.49 kJ = − 3. Find the efficiency of this cycle. and (3) heating at constant volume to its original pressure. as expected because the system returns to its initial state.75 (b) The efficiency of the cycle is given by: Substitute numerical values and evaluate ε: ε= Wby Qin = − W23 + (− W41 ) Q12 + Q23 ε= 4. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.0 L ) = −35.639 atm (1) ε= W Qh No heat enters or leaves the system during the adiabatic expansion: Find the heat entering or leaving the system during the isobaric compression: Find the heat entering or leaving the system during the constantvolume process: Apply the 1st law of thermodynamics to the cycle ( ΔEint. cycle = 0 ) to obtain: Q12 = 0 7 Q23 = CP ΔT23 = 7 2 RΔT23 = 2 PΔV23 = 7 2 (1.0 L ⎠ 1.1800 Chapter 19 Picture the Problem The three steps in the process are shown on the PV diagram.0 20.0 L ) = 41.0 atm ⋅ L + 41.00 atm )(10.0 atm ⋅ L Substitute numerical values in equation (1) and evaluate ε : ε= 6.0 V(L) The pressures and volumes at the end points of the adiabatic expansion are related according to: Substitute numerical values and evaluate P1: Express the efficiency of the cycle: ⎛ V2 ⎞ P1V1 = P2V2 ⇒ P1 = ⎜ ⎟ P2 ⎜V ⎟ ⎝ 1⎠ γ γ γ ⎛ 20.4 (1. P (atm) 2.0 atm ⋅ L Q31 = C V ΔT31 = 5 RΔT31 = 5 ΔPV31 2 2 = 5 2 (2.639 atm − 1.0 L ⎞ P1 = ⎜ ⎟ ⎝ 10.0 atm ⋅ L = 15% 41atm ⋅ L .639 2 1 1 3 2 0 0 10.0 atm ⋅ L = 6.0 atm ⋅ L Won = ΔEint − Qin = −Qin = Q12 + Q23 + Q31 = 0 − 35.0 L − 20.00 atm) = 2.00 atm )(10. and (4) heating at constant volume to its original temperature of 400 K.1 + Qh. 4 W + W3 W = 1 Qh Qh. 2 + Qh. B. 3. and 4 represent the four steps through which the gas is taken.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume. and D identify the four states of the gas and the numerals 1.00 mol of an ideal gas initially at a volume of 24. A. Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle. (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume.5 1 0.1 + Qh. Sketch the cycle on a PV diagram and find its efficiency. 2. W2 = W4 = 0: Because the internal energy of the gas increases in step 4 while no work is done.0 J/K. 2 + Qh.5 0 0 10 20 4 D A 1 3 C B 2 400 K 300 K 30 40 V (L) 50 60 Express the efficiency of the cycle: ε= W1 + W2 + W3 + W4 W = Qh Qh. 2 P (atm) 1.The Second Law of Thermodynamics 1801 33 •• An engine using 1. C.3 + Qh. The PV diagram of the cycle is shown to the right. heat enters the system only during these processes: The work done during the isothermal expansion (1) is given by: The work done during the isothermal compression (3) is given by: ε= ε= ⎛ VB W1 = nRT ln⎜ ⎜V ⎝ A ⎞ ⎟ ⎟ ⎠ ⎛ VD ⎞ W3 = nRTc ln⎜ ⎜V ⎟ ⎟ ⎝ C⎠ .3 + Qh. 4 (1) Because steps 2 and 4 are constantvolume processes. and because the internal energy does not change during step 1 while work is done by the gas.1 + Qh. 4 W1 + 0 + W3 + 0 W = Qh Qh. Assume that Cv = 21. 00 mol)⎜ 8.0 K (400 K − 300 K ) 400 K + J ⎞ ⎛ (1. the heat that enters the system during this isothermal expansion is given by: The heat that enters the system during the constant-volume step 4 is given by: Substituting in equation (1) yields: ⎛ VB Q1 = W1 = nRTh ln⎜ ⎜V ⎝ A ⎞ ⎟ ⎟ ⎠ Q4 = C V ΔT = C V (Th − Tc ) ⎛ VD ⎞ ⎛ VB ⎞ nRTh ln⎜ ⎜V ⎟ ⎟ + nRTc ln⎜ ⎜V ⎟ ⎟ A ⎠ ⎝ ⎝ C⎠ ε= ⎛ VB ⎞ nRTh ln⎜ ⎜V ⎟ ⎟ + C V (Th − Tc ) ⎝ A⎠ Noting the VB V 1 = 2 and D = .00 mol of an ideal monatomic gas initially at a volume of 25. substitute and simplify to obtain: VA VC 2 ⎛1⎞ Th ln (2 ) + Tc ln⎜ ⎟ Th − Tc ⎝ 2 ⎠ = Th ln (2) − Tc ln (2) = ε= C CV C (Th − Tc ) Th ln (2) + V (Th − Tc ) Th ln (2) + V (Th − Tc ) Th + nR nR nR ln (2) Substitute numerical values and evaluate ε: ε= 400 K − 300 K = 13. Because the change in internal energy is zero for the isothermal process.1% J 21.314 ⎟ ln (2) mol ⋅ K ⎠ ⎝ 34 •• Figure 19-15 shows the cycle followed by 1. we can use the expression for the work done on or by a gas during an isothermal process . All the processes are quasi-static. and (c) the efficiency of the cycle.1802 Chapter 19 Because there is no change in the internal energy of the system during step 1. (b) the heat transfer for each part of the cycle. Determine (a) the temperature of each numbered state of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes.0 L. 24 kJ = 0.96 kJ because.24 kJ 2 2⎜ mol ⋅ K ⎠ ⎝ (c) Express the efficiency of the cycle: Apply the 1st law of thermodynamics to the cycle: ε= W W = Qin Q12 + Q23 (1) W = ∑ Q = Q12 + Q23 + Q31 = 3.0 L ⎟ ⎟ = 3. ΔEint = 0 .46 kJ ⎜V ⎟ ⎟ = (1. (a) Use the ideal-gas law to find the temperature at point 1: T1 = P1V1 = nR (100 kPa )(25.00 mol)⎜ 8.314 ⎝ J ⎞ ⎟ mol ⋅ K ⎠ = 301 K Use the ideal-gas law to find the temperatures at points 2 and 3: T2 = T3 = P2V2 nR (200 kPa )(25.0 L ) (1. we can find the efficiency of the cycle from its definition.314 ⎟ (301 K − 601 K ) = − 6.74 kJ + 3. Finally.46 kJ − 6.00 mol)⎛ ⎜ 8.0 L ) = J ⎞ (1. . for the cycle.74 kJ 2 2⎜ mol ⋅ K ⎠ ⎝ Find the heat entering or leaving the system for the isothermal process from 2 → 3: ⎛ V3 ⎞ ⎛ 50.00 mol)⎛ ⎜ 8.314 ⎟ mol ⋅ K ⎠ ⎝ = 601 K (b) Find the heat entering the system for the constant-volume process from 1 → 2: J ⎞ ⎛ Q12 = C V ΔT12 = 3 RΔT12 = 3 8.The Second Law of Thermodynamics 1803 to find the heat flow during such a process.0 L ⎞ J ⎞ ⎛ Q23 = nRT2 ln⎜ ⎜ 25.314 ⎟ (601 K − 301 K ) = 3.314 mol ⋅ K ⎟ (601 K )ln⎜ ⎝ ⎠ ⎝ ⎠ ⎝ 2⎠ Find the heat leaving the system during the isobaric compression from 3 → 1: J ⎞ ⎛ Q31 = C P ΔT31 = 5 RΔT31 = 5 8. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin.1804 Chapter 19 Substitute numerical values in equation (1) and evaluate ε : ε= 0.46 kJ 35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1: Substitute numerical values and evaluate T2: Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain: Substitute numerical values and evaluate T3: Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain: Substitute numerical values and evaluate T4: (b) The efficiency of the cycle is: PV P P PV 1V1 = 2 2 ⇒ T2 = T1 2 2 = T1 2 T1 T2 P P 1V1 1 T2 = (200 K ) (3. We can find the efficiency of the cycle from its definition.74 kJ + 3.0 atm ) = (1.0 atm − 1.0 atm ) 600 K ε= W Qin (1) Use the area of the rectangle to find the work done each cycle: W = ΔPΔV = (300 L − 100 L )(3.0 atm ) 600 K T3 = T2 P3V3 V = T2 3 P2V2 V2 T3 = (600 K ) (300 L ) = (100 L ) 1800 K T4 = T3 P4V4 P = T3 4 P3V3 P3 T4 = (1800 K ) (1.0 atm ) = 400 atm ⋅ L .96 kJ = 13% 3. The temperature of state 1 is 200 K.0 atm ) = (3. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle. known as the Stirling engine has been promoted as a means of producing power from solar energy. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. Picture the Problem (a) The PV and ST cycles are shown below. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume.0 atm )(100 L ) = T1 200 K = 0. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. and (4) cooling of the gas at constant volume. express Qin: Qin = Q12 + Q23 = C V ΔT12 + C P ΔT23 =5 nRΔT12 + 7 nRΔT23 2 2 = (5 ΔT12 + 7 ΔT23 )nR 2 2 Substitute numerical values and evaluate Qin: L ⋅ atm ⎞ ⎛ 7 Qin = [5 1800 K − 600 K )]⎜ 0. (a) Sketch PV and ST diagrams for the Stirling cycle.50 L ⋅ atm/K Noting that heat enters the system between states 1 and 2 and states 2 and 3. P S 2 (2) (3) Th (1) Tc (4) ΔV = 0 3 Th Tc 4 V ΔV = 0 1 T . an old design for a heat engine. (3) an isothermal expansion of the gas.The Second Law of Thermodynamics 1805 Apply the ideal-gas law to state 1 to find the product of n and R: nR = P1V1 (1.50 ⎟ = 2600 atm ⋅ L 2 (600 K − 200 K ) + 2 ( K ⎠ ⎝ Substitute numerical values in equation (1) and evaluate ε : ε= 400 atm ⋅ L = 15% 2600 atm ⋅ L 36 ••• Recently. (a) Calculate the efficiency of a heat engine operating between body temperature (98. give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies. Life’s Devices. Nature has never evolved a heat engine″—Steven Vogel. . Princeton University Press (1988). and compare this to the human body’s efficiency for converting chemical energy into work (approximately 20%).1806 Chapter 19 (b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle: ΔS cycle = ΔS12 + ΔS 23 + ΔS 34 + ΔS 41 (1) Express the entropy change for the isothermal process from state 1 to state 2: Similarly.6ºF) and a typical outdoor temperature (70ºF). ⎛ V1 ⎞ ⎛ V2 ⎞ ΔS 34 = nR ln⎜ ⎜V ⎟ ⎟ = − nR ln⎜ ⎜V ⎟ ⎟ ⎝ 2⎠ ⎝ 1⎠ The change in entropy for a constantvolume process is given by: ΔS isochoric dQ f nCV dT =∫ =∫ T T Ti ⎛ Tf = nC V ln⎜ ⎜T ⎝ i ⎞ ⎟ ⎟ ⎠ T For the constant-volume process from state 2 to state 3: For the constant-volume process from state 4 to state 1: Substituting in equation (1) yields: ⎛ Tc ΔS 23 = C V ln⎜ ⎜T ⎝ h ⎛ Th ΔS 41 = C V ln⎜ ⎜T ⎝ c ⎞ ⎟ ⎟ ⎠ ⎞ ⎛ Tc ⎟ ⎜T ⎟ = −C V ln⎜ ⎝ h ⎠ ⎞ ⎟ ⎟ ⎠ ⎛ V2 ⎞ ⎛ Tc ⎞ ⎛ V2 ⎞ ⎛ Tc ⎞ ΔS cycle = nR ln⎜ ⎜V ⎟ ⎟ + CV ln⎜ ⎜T ⎟ ⎟ − nR ln⎜ ⎜V ⎟ ⎟ − CV ln⎜ ⎜T ⎟ ⎟= 0 ⎝ 1⎠ ⎝ h⎠ ⎝ 1⎠ ⎝ h⎠ 37 •• ″As far as we know. Does this efficiency comparison contradict the second law of thermodynamics? (b) From the result of Part (a). because V2 = V3 and V1 = V4. the entropy change for the isothermal process from state 3 to state 4 is: ⎛ V2 ⎞ ΔS12 = nR ln⎜ ⎜V ⎟ ⎟ ⎝ 1⎠ ⎛ V4 ⎞ ΔS 34 = nR ln⎜ ⎜V ⎟ ⎟ ⎝ 3⎠ or. and a general knowledge of the conditions under which most warm-blooded organisms exist. internal body temperatures would have to be maintained at an unreasonably high level.16% 310 K The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. Picture the Problem The working fluid will be modeled as an ideal gas and the process will be modeled as quasistatic. Express the efficiency of the cycle in terms of Qc and Qh: Express Q for the isobaric warming process bc: ε= Q W Qh − Qc = = 1− c Qh Qh Qh Qbc = Qh = C P (Tc − Tb ) . Find the efficiency of this cycle in terms of the volumes Va. Heat enters the system during the isobaric process bc and leaves the system during the isovolumetric process da.The Second Law of Thermodynamics 1807 Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures. To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. we eat food to get the energy that we need. process cd is an adiabatic expansion. To make a heat engine work with appreciable efficiency. Note that no heat enters or leaves the system during the adiabatic processes ab and cd. Rather. Process ab is an adiabatic compression. process bc is an expansion at constant pressure. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that we don’t get our energy from heat swapping between our body and the environment. Vb and Vc. 38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a diesel engine. (b) Most warm-blooded animals survive under roughly the same conditions as humans. and process da is cooling at constant volume. (a) Express the maximum efficiency of an engine operating between body temperature and 70°F: Use T = 5 9 εC = 1 − Tc Th (tF − 32) + 273 to obtain: Tbody = 310 K and Troom = 294 K Substitute numerical values and evaluate ε C : ε C = 1− 294 K = 5. relate the temperatures Ta and Tb to the volumes Va and Vb: Proceeding similarly.1808 Chapter 19 Because CV is independent of T. apply the idealgas law to relate Tb and Tc: Tb Vb = Tc Vc ⎛ Substitute for the ratio of Tb to Tc and simplify to obtain: ⎛ Vc ⎜ ⎜V ε = 1− ⎝ a ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Vb ⎜ ⎜V ⎝ a ⎛ Vb ⎞ γ⎜ ⎟ ⎜1 − ⎟ ⎝ Vc ⎠ V − b Vc γ ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Vc Vc ⎜ ⎜V Va ⋅ = 1− ⎝ a Vc Va ⎞ ⎟ ⎟ ⎠ ⎛ Vb ⎜ ⎜V ⎝ a ⎛ Vc Vb ⎞ γ⎜ ⎜V − V ⎟ ⎟ a ⎠ ⎝ a V − b Vc γ −1 ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Vc ⎞ ⎛ Vb ⎞ ⎜ ⎜V ⎟ ⎟ −⎜ ⎜V ⎟ ⎟ Vcγ − Vbγ a ⎠ a ⎠ ⎝ ⎝ = 1− = 1 − γ −1 γVa (Vc − Vb ) ⎛ Vc Vb ⎞ γ⎜ ⎜V − V ⎟ ⎟ a ⎠ ⎝ a γ . relate the temperatures Tc and Td to the volumes Vc and Vd: Use equations (1) and (2) to eliminate Ta and Td: Qda = Qc = C V (Td − Ta ) ε = 1− (T − Ta ) C V (Td − Ta ) = 1− d C P (Tc − Tb ) γ (Tc − Tb ) Vbγ −1 (1) Vaγ −1 TaVaγ −1 = TbVbγ −1 ⇒ Ta = Tb TcVcγ −1 = TdVdγ −1 ⇒ Td = Tc Vcγ −1 (2) Vdγ −1 ⎛ Vcγ −1 Vbγ −1 ⎞ ⎜ T T − ⎜ c V γ −1 b V γ −1 ⎟ ⎟ d a ⎝ ⎠ ε = 1− γ (Tc − Tb ) ⎛⎛ V ⎜⎜ c ⎜⎜ V ⎝⎝ a ⎞ ⎟ ⎟ ⎠ γ −1 Because Va = Vd: ε = 1− T ⎛ Vb − b⎜ Tc ⎜ ⎝ Va Tb ⎞ ⎟ Tc ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ γ −1 ⎞ ⎟ ⎟ ⎠ γ⎜ ⎜1 − ⎝ Noting that Pb = Pc. Qda (the constant-volume cooling process) is given by: Substitute for Qh and Qc and simplify using γ = C P C V to obtain: Using an equation for a quasistatic adiabatic process. Hot reservoir at temperature Th ⇓ Ordinary refrigerator 300 J 300 J 500 J (a) Cold reservoir at temperature Tc (b) (c ) If two curves that represent quasi-static adiabatic processes could 40 •• intersect on a PV diagram. Show that such a cycle violates the second law of thermodynamics. we would completely convert heat to mechanical energy. where we assume that we start with the isothermal expansion.This violates the refrigerator statement of the second law.The Second Law of Thermodynamics 1809 Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Then. a cycle could be completed by an isothermal path between the two adiabatic curves shown in Figure 19-18. in violation of the second law of thermodynamics. Determine the Concept The work done by the system is the area enclosed by the cycle. and show how a perfect engine working with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram). ⇓ 500 J Perfect heat engine Perfect refrigerator ⇓ ⇓ 800 J 300 J 500 J ⇓ ⇓ ⇓ . Suppose the heat-engine statement of the second law is false. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Then a ″perfect″ heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. It is only in this expansion that heat is extracted from a reservoir. Thus. Assume that the heat-engine statement of the second law of thermodynamics is false. the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator. without exhausting any heat to a cold reservoir. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). There is no heat transfer in the adiabatic expansion or compression. where εC is the Carnot efficiency. (a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs: (b) Using the definition of efficiency.0 W 33.3 J Qc = Qh − W = 100 J − 33. We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle.7 J = = 2. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from W = ε CQh . . express and evaluate the refrigerator’s coefficient of performance: ε C = 1− Tc 200 K = 1− = 33. We can find the COP of the engine working as a refrigerator from its definition.1810 Chapter 19 Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. relate the work done each cycle to the heat absorbed from the hot reservoir: (c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done: (d) Using its definition.3% Th 300 K W = ε C Qh = (0.7 J = 67 J COP = Qc 66.333)(100 J ) = 33.3 J An engine absorbs 250 J of heat per cycle from a reservoir at 300 K 42 • and releases 200 J of heat per cycle to a reservoir at 200 K.3 J = 66. how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using ε = 1 − Tc / Th and the work done per cycle from ε = W / Qh . (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle. deliver 200 J to the hot reservoir. A second engine working between the same two reservoirs also does 100 J of work per cycle. Working as a heat engine.3 J ⎝ ⎠ ΔW = 83. in violation of the second law. is greater than 30%. is 140 J/(1 − ε2) > 200 J. 44 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 20%. and require 60 J of energy to operate. Show that if the second engine has an efficiency greater than 30%. the two engines working together would violate the heat-engine statement of the second law.The Second Law of Thermodynamics (a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the lowtemperature reservoir: (b) Express the additional work done if the engine is reversible: Relate the work done by a reversible engine to its Carnot efficiency: Substitute numerical values and evaluate W: Substitute numerical values in equation (1) and evaluate ΔW: 1811 ε= Q W Qh − Qc = = 1− c Qh Qh Qh 200 J = 20. The end result of all this is that the second engine can run the refrigerator. Then it will remove 140 J from the cold reservoir. and do additional mechanical work. A second engine working between the same two reservoirs also releases 140 J per cycle to the cold reservoir. Determine the Concept Let the first engine be run as a refrigerator. Now take the second engine and run it between the same reservoirs. Show that if the efficiency of the second engine is greater . replacing the heat taken from the cold reservoir. Working as a heat engine. the efficiency of this engine. the heat removed from the hot reservoir by this engine. and let it eject 140 J into the cold reservoir. it does 100 J of work per cycle.3 J − 50 J = 33 J 43 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 30%. it releases 140 J per cycle of heat to the cold reservoir. If ε2. and the work done by this engine is W = ε2Qh2 > 60 J. thus replacing the heat removed by the refrigerator. then Qh2.0% 250 J (1) = 1− ΔW = WCarnot − WPart (a ) ⎛ Tc W = ε C Qh = ⎜ ⎜1 − T h ⎝ ⎞ ⎟ ⎟Qh ⎠ ⎛ 200 K ⎞ W =⎜ ⎜1 − 300 K ⎟ ⎟ (250 J ) = 83. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir. it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir.1812 Chapter 19 than 20%. then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. 45 •• A Carnot engine works between two heat reservoirs as a refrigerator. The net result is then that no net work is done by the two systems working together.00. Now let the second engine. but a finite amount of heat is transferred from the cold reservoir to the hot reservoir. this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. So running the engine described in Part (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law.2.2. (a) What is the efficiency of the Carnot engine when it works as a heat engine between the same two reservoirs? (b) Show that no other engine working as a refrigerator between the same two reservoirs can have a COP greater than 2. in violation of the refrigerator statement of the second law. 46 •• A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 77.0 K. Determine the Concept If the reversible engine is run as a refrigerator. the two engines working together would violate the refrigerator statement of the second law. During each cycle. with ε2 > 0. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle. 100 J of heat are absorbed and 150 J are released to the hot reservoir. (a) The efficiency of the Carnot engine is given by: εC = W 50 J = = 33% Qh 150 J (b) If the COP > 2. Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs. Because ε2 > 0. operate between the same two heat reservoirs and use it to drive the refrigerator. how much work does it do? (c) How much heat does it release to the low-temperature reservoir during each cycle? (d) What is the coefficient of performance of this engine when it works as a refrigerator between these two reservoirs? . 3% Th 300 K W = ε C Qh = (0.3 J Qc = Qh − W = 100 J − 74.3 J COP = 47 •• [SSM] In the cycle shown in Figure 19-19.00 atm.0ºC. Once we’ve calculated these quantities.35 W 74. volumes. .3 J = 26 J Qc 26 J = = 0. The gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically until its pressure is again 1. we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures. Find (a) the temperature after the adiabatic expansion.743)(100 J ) = 74. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures. 1. We can use Q = C V ΔT and Q = C P ΔT to find the heat entering and leaving during the constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle.00 mol of an ideal diatomic gas is initially at a pressure of 1. (c) the efficiency of this cycle.The Second Law of Thermodynamics 1813 Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities.00 atm and a temperature of 0. (b) the heat absorbed or released by the system during each step. The work done each cycle by the Carnot engine is given by W = ε CQh and we can use the conservation of energy to find the heat rejected to the low-temperature reservoir.0 K = 1− = 74. (a) The efficiency of a Carnot engine depends on the temperatures of the hot and cold reservoirs: (b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir: (c) Apply conservation of energy to obtain: (d) Using its definition. and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. It is then compressed at constant pressure back to its original state. express and evaluate the refrigerator’s coefficient of performance: ε C = 1− Tc 77. and pressures at the end points of each process in the given cycle. V T3 = T1 3 V1 PV T P PV 1V1 = 2 2 ⇒ P2 = 1 1 2 T1 T2 V2T1 (1) Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1: Because V1 = V2: P2 = P1 T2 423 K = (1.00 atm ) = 1. because P1 = P3.6 L = 373 K 22.4 L and t 3 = T3 − 273 = 100°C (b) Process 1→2 takes place at constant volume (note that γ = 1.1814 Chapter 19 (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1: P PV 1V1 = 3 3 T1 T3 or.314 ⎟ (273 K − 373 K ) 2⎜ mol ⋅ K ⎠ ⎝ = − 2.4 V3 = (22.91 kJ .55 atm T1 273 K 1 Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3: Noting that V1 = 22. evaluate V3: ⎛P ⎞γ γ γ 1 ⎜ ⎟ V V ⇒ = P V = P V 3 1⎜ 1 1 3 3 ⎟ ⎝ P3 ⎠ 1 ⎛ 1.314 ⎟ (423 K − 273 K ) 2⎜ mol ⋅ K ⎠ ⎝ = 3.6 L ⎝ ⎠ Substitute numerical values in equation (1) and evaluate T3 and t3: T3 = (273 K ) 30.4 corresponds to a diatomic gas and that CP – CV = R): Process 2→3 takes place adiabatically: Process 3→1 is isobaric (note that CP = CV + R): Q12 = CV ΔT12 = 5 RΔT12 2 J ⎞ ⎛ =5 8.55 atm ⎞ 1.4 L.12 kJ Q23 = 0 Q31 = C P ΔT31 = 7 RΔT12 2 J ⎞ ⎛ =7 8.4 L )⎜ ⎜ 1atm ⎟ ⎟ = 30. 0%. (a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by: The efficiency of a Carnot engine operating between temperatures Fc and Th is: ε steam engine 0. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.0ºC.12 kJ + 0 − 2.91 kJ = 0.00 h? Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50. how much heat does the engine release to its surroundings in 1.300 = ε max ε max ε max = 1 − Tc 323 K = 1− = 40.21 kJ Substitute numerical values in equation (2) and evaluate ε : (d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K: ε= 0.12 kJ Tc 273 K = 1− = 35. Evaluating W yields: W = ∑ Q = Q12 + Q23 + Q31 = 3. because ΔEint.The Second Law of Thermodynamics (c) The efficiency of the cycle is given by: Apply the first law of thermodynamics to the cycle: 1815 ε= W Qin (2) ΔEint = Qin + Won or.7% 3. cycle = 0 (the system begins and ends in the same state) and Won = −Wby the gas = Qin .52% Th 543 K . (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW.00 h.21kJ = 6. Your team has measured its efficiency to be 30.5% Th 423 K εC = 1− 48 •• You are part of a team that is completing a mechanical-engineering project. 1816 Chapter 19 Substituting for εmax yields: ε steam engine 0. What is the minimum power of the electric motor when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. The house is located where.300 = = 74. We can apply the definition of power to find the minimum power needed to run the heat pump. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must the minimum power of the electric motor driving the heat pump be? (c) In reality. The temperature of the air in the air handler inside the house is to be 40ºC. relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle: Substitute for Qh in the expression for Qc and simplify to obtain: ε= W Qh − Qc = ⇒ Qc = (1 − ε )Qh Qh Qh PΔt Qh = W ε = ε Qc = (1 − ε ) PΔt ε ⎛1 ⎞ = ⎜ − 1⎟ PΔt ⎝ε ⎠ Substitute numerical values and evaluate Qc (1. .740ε max (b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine: Using its definition. the COP of the heat pump will be only 60 percent of the ideal value.300 ⎠ ⎝ *Heat Pumps 49 • [SSM] As an engineer.00 h ) : kJ ⎞ ⎞⎛ ⎛ 1 Qc (1.68 GJ s ⎠ ⎝ 0.4052 or ε steam engine = 0. in January.05% ε max 0. you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house.00 h ) = ⎜ − 1⎟ ⎜ 200 ⎟ (3600 s ) = 1. the average outside temperature is –10ºC. 26 313 K − 263 K = 6.The Second Law of Thermodynamics (a) Express the COPHP in terms of Th and Tc: COPHP = = Qh Qh = W Qh − Qc Th 1 1 = = Q T Th − Tc 1− c 1− c Qh Th 1817 Substitute numerical values and evaluate COPHP: COPHP = 313 K = 6.00 min under these conditions? Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time.max ) ε HP where ε HP is the efficiency of the heat pump.26) 50 • A refrigerator is rated at 370 W. Th.60)(6.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible refrigerator.3 (b) The COPHP is also given by: COPHP = Pout Pout ⇒ Pmotor = COPHP Pmotor Substitute numerical values and evaluate Pmotor: (c) The minimum power of the electric motor is given by: Pmotor = 20 kW = 3. and Δt.26 dQc dt ε (COPHP. P.2 kW 6. (a) What is the maximum amount of heat it can absorb from the food compartment in 1. how much heat can it absorb from the food compartment in 1.00 min if the foodcompartment temperature of the refrigerator is 0. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc.3 kW (0. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: Qc = (COP )W = (COP )PΔt .0ºC and it releases heat into a room at 20. Pmin dQc = dt = Substitute numerical values and evaluate Pmin: Pmin = 20 kW = 5. Th.30 MJ min ⎠ ⎝ 293 K − 273 K ⎠ ⎝ (b) If the COP is 70% of the efficiency of an ideal pump: ' Qc = (0. Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: Qc = (COP )W = (COP )PΔt .70 )(303 kJ ) = 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of the refrigerator is 70% of that of a reversible pump.21 MJ 51 • A refrigerator is rated at 370 W.00 min × ⎟ = 303 kJ = 0. how much heat can it absorb from the food compartment in 1.00 min if the temperature in the compartment is 0.1818 Chapter 19 Express the COP in terms of Th and Tc and simplify to obtain: COP = = Qc Q Q −W = c = h W ε Qh ε Qh 1− ε ε = 1 ε −1 = 1 −1 Tc 1− Th = Tc Th − Tc Substituting for COP yields: ⎛ Tc ⎞ Qc = ⎜ ⎜T −T ⎟ ⎟ PΔt ⎝ h c⎠ Substitute numerical values and evaluate Qc: 273 K 60 s ⎞ ⎛ ⎞ ⎛ Qc = ⎜ ⎟ (370 W )⎜1.00 min? Is the COP for the refrigerator greater when the temperature of the room is 35ºC or 20ºC? Explain. P. (a) What is the maximum amount of heat it can absorb for the food compartment in 1. and Δt. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc. ceiling.70 )(173 kJ ) = 0. whose COP is half the COP of a reversible heat pump. and the temperature at the air handler in the room is 112°F. ceilings and floors. Also assume that the heat capacity of the floor.50 m.The Second Law of Thermodynamics 1819 Express the COP in terms of Th and Tc and simplify to obtain: COP = = = Qc Q Q −W = c = h W εQh εQh 1− ε ε = 1 ε −1 Tc 1 −1 = T Th − Tc 1− c Th Substituting for COP yields: ⎛ Tc ⎞ Qc = ⎜ ⎜T −T ⎟ ⎟ PΔt ⎝ h c⎠ Substitute numerical values and evaluate Qc: 273 K 60 s ⎞ ⎛ ⎞ ⎛ Qc = ⎜ ⎟ = 173 kJ = 0. Your bedroom’s dimensions are 5. If the pump’s electric power consumption is 750 W.50 m × 2. walls. how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1.12 MJ Because the temperature difference increases when the room is warmer.00 m × 3.17 MJ ⎟ (370 W )⎜1. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. The air temperature should increase from 63°F to 68°F. The outside temperature is 35°F .005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows. the COP decreases.00 min × min ⎠ ⎝ 308 K − 273 K ⎠ ⎝ (b) If the COP is 70% of the efficiency of an ideal pump: ' Qc = (0. 52 ••• You are installing a heat pump. . walls and furniture are negligible. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the pump’s power consumption to find your waiting time. 50 m × 2. COPHP = We’re given that the coefficient of performance of the heat pump is half the coefficient of performance of an ideal heat pump: COPHP = Qh 1 = COPmax W 2 ⎛ Th ⎞ ⎜ ⎟ =1 2⎜ ⎟ − T T ⎝ h c⎠ 2Qh Substituting for COPHP yields: Δt = ⎛ Th ⎞ ⎜ ⎜ T −T ⎟ ⎟P ⎝ h c⎠ The heat required to warm the room is related to the volume of the room. Δt = 2 ρVcΔT ⎛ Th ⎞ ⎜ ⎜ T −T ⎟ ⎟P ⎝ h c⎠ Substitute numerical values and evaluate Δt: ⎛ kg ⎞ J ⎞⎛ 5 C° ⎞ ⎛ ⎟ 2⎜1.50 m )⎜ 1005 ⎜ 5 F° × ⎟ ⎜ ⎟ kg ⋅ C° ⎠ ⎝ 9 F° ⎠ m ⎠ ⎝ ⎝ Δt = = 56 s 317 K ⎛ ⎞ ⎜ ⎟ (750 W ) ⎝ 317 K − 275 K ⎠ Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. P is the power consumption of the heat pump.293 3 ⎟ (5. the density of air.1820 Chapter 19 The coefficient of performance of the heat pump is defined as: Qh Q Qh = h ⇒ Δt = (COPHP )P W PΔt where Qh is the heat required to raise the temperature of your bedroom. You return just in time to see the last drop converted into steam. and Δt is the time required to warm the bedroom. and the desired increase in temperature: Substitute for Qh to obtain: Qh = mcΔT = ρVcΔT where ρ is the density of air and c is its specific heat capacity. What is the change in entropy of the water associated with its change of state from liquid to gas? . The pan originally held 1.00 m × 3.00 L of boiling water. Because heat is removed from liquid water when it freezes.05 K 54 • What is the change in entropy of 1.00 L )⎜ ⎜ L⎠ kg ⎟ ⎝ ⎝ ⎠ = 373 K kJ = 6. See Appendix C for the molar mass of water and Table 18-2 for the latent heat of fusion of water. Qremoved = − nM H 2 O Lf from H 2 O . The change in entropy of the water is given by:: The heat removed from the water as it freezes is the product of its mass and latent heat of fusion: ΔS H 2 O = Qremoved T from H 2 O Qremoved = −mLf from H 2 O or. The change in entropy of the water is given by: ΔS H 2 O = Qabsorbed by H 2 O T The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: Substituting for Qabsorbed yields: by H 2 O Qabsorbed = mLv = ρVLv by H 2 O ΔS H 2 O = ρVLv T Substitute numerical values and evaluate ΔS H 2O : ΔS H 2 O ⎛ kg ⎞ kJ ⎞ ⎛ 2257 ⎟ ⎜1.0°C? Picture the Problem We can use the definition of entropy change to find the change in entropy of the liquid water as it freezes.00 ⎟ (1. the change in entropy of the liquid water is negative. See Table 18-2 T for the latent heat of vaporization of water.0ºC that freezes to ice at 0.00 mol of liquid water at 0.The Second Law of Thermodynamics 1821 Picture the Problem Because the water absorbed heat in the vaporization process Qabsorbed by H 2 O its change in entropy is positive and given by ΔS H 2 O = . because m = nM H 2 O . 015 333. The change in entropy of the universe resulting from this freezing and cooling process is given by: Express ΔS water : Express ΔS freezing : ΔS u = ΔS water + ΔS freezer (1) ΔS water = ΔS freezing + ΔS cooling ΔS freezing = − Qfreezing Tfreezing (2) (3) where the minus sign is a consequence of the fact that energy is leaving the water as it freezes. Note that. initially liquid at 0. The water. the net entropy of the universe increases.1822 Chapter 19 Substitute numerical values and evaluate ΔS H 2O : g ⎞⎛ J⎞ ⎛ − (1.0ºC.5 ⎟ ⎟⎜ ⎜ mol ⎠ ⎝ g⎟ ⎝ ⎠ = − 22. while the entropy of the water decreases. Relate Qfreezing to the latent heat of fusion and the mass of the water: Substitute in equation (3) to obtain: ΔS freezing = − mLf Tfreezing ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ Qfreezing = mLf Express ΔS cooling : ⎛ Tf ΔS cooling = mC p ln⎜ ⎜T ⎝ i ΔS water = Substitute in equation (2) to obtain: ⎛ Tf − mLf + mCp ln⎜ ⎜T Tfreezing ⎝ i . is frozen into ice and cooled to –10ºC.00 mol)⎜18.0 J = K 273 K ΔS H 2O 55 •• Consider the freezing of 50. Show that even though the entropy of the water decreases. the entropy of the freezer increases.0 g of water once it is placed in the freezer compartment of a refrigerator. Picture the Problem The change in the entropy of the universe resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Assume the walls of the freezer are maintained at –10ºC. 00 mol of an ideal gas at 400 K expand quasistatically and isothermally from an initial volume of 40. the entropy of the universe increases. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes.40 J/K + ⎥ 263 K ⎥ ⎦ and.0 L. (a) The entropy change of the gas is given by: ΔS gas = Q T (1) .0500 kg ) ⎢− ⎜ ⎜ ⎟ kg ⋅ K ⎟ 273 K ⎢ ⎝ ⎠ ⎝ 273 K ⎠ ⎢ ⎣ ⎤ J ⎞ J ⎛ ⎟ ( ) +⎜ 273 K − 263 K 2100 ⎥ kg ⋅ K ⎟ kg ⎜ ⎝ ⎠ ⎥ = 2.5 ×10 kg ⎛ J ⎞ ⎛ 263 K ⎞ ⎟ ⎟ ln⎜ +⎜ 2100 ΔS u = (0.The Second Law of Thermodynamics 1823 Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice.5 ×10 3 56 • In this problem. express ΔS freezer : ΔS freezer = = ΔQice ΔQcooling ice + Tfreezer Tfreezer mCp ΔT mLf + Tfreezer Tfreezer Substitute for ΔS water and ΔS freezer in equation (1): ΔS u = ⎛ Tf − mLf + mCp ln⎜ ⎜T Tfreezing ⎝ i mCp ΔT ⎞ mLf ⎟ + + ⎟ T Tfreezer ⎠ freezer ⎞ Lf + C p ΔT ⎤ ⎟ ⎥ ⎟+ T ⎥ freezer ⎠ ⎦ ⎡ − Lf ⎛ Tf = m⎢ + C p ln⎜ ⎜T ⎢ Tfreezing ⎝ i ⎣ Substitute numerical values and evaluate ΔSu: ⎡ 3 J ⎢ 333. 2. because ΔSu > 0.0 L to a final volume of 80. 333. and during step 5 it absorbs heat from a reservoir at temperature T3.00 mol)⎜ 8. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K. Q = −Won ⎛ Vi Won = nRT ln⎜ ⎜V ⎝ f ⎛ Vi ⎞ ⎜ ⎟ = − ⇒ ln Q nRT ⎜V ⎟ ⎝ f ⎠ ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ The work done on the gas is given by: Substitute for Q in equation (1) to obtain: ⎛ Vi ΔS gas = −nR ln⎜ ⎜V ⎝ f Substitute numerical values and evaluate ΔS: J J ⎞ ⎛ 40.1824 Chapter 19 Apply the first law of thermodynamics to the isothermal process to express Q in terms of Won: Q = ΔEint − Won or. (a) Because S is a state function of the system. during step 3 the system absorbs 200 J of heat from a reservoir at 400 K. Because the entropy change for the complete cycle is the sum of the entropy changes for each process.) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible. because ΔEint = 0 for an isothermal expansion of a gas. and because the system’s final state is identical to its initial state: ΔS system = 0 1 complete cycle .5 J/K. we can find the temperature T3 from the entropy changes during the 1st two processes and the heat released during the third.5 ΔS gas = −(2.0 L ⎠ ⎝ (b) Because the process is reversible: ΔS u = 0 Remarks: The entropy change of the environment of the gas is −11. 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next.314 ⎟ ln⎜ ⎜ ⎟ K mol ⋅ K ⎠ ⎝ 80. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps. during which the total work done by the system is 100 J. what is the temperature T3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. (During steps 2.0 L ⎞ ⎛ ⎟ = 11. The Second Law of Thermodynamics 1825 (b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system: Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir: Solving for T3 yields: ΔS1 + ΔS 2 + ΔS 3 + ΔS system = 0 or Q1 Q2 Q3 + + +0 = 0 T1 T2 T3 − 300 J − 200 J 400 J + + =0 300 K 400 K T3 T3 = 267 K 58 •• In this problem, 2.00 mol of an ideal gas initially has a temperature of 400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. What is (a) the entropy change of the gas and (b) the entropy change of the universe? Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change ΔS for a free expansion from Vi to Vf is the same as ΔS for an isothermal process from Vi to Vf. We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS for the ideal gas as a function of its initial and final volumes. (a) The entropy change of the gas is given by: Apply the first law of thermodynamics to the isothermal process to express Q: ΔS gas = Q T (1) Q = ΔEint − Won or, because ΔEint = 0 for a free expansion of a gas, Q = −Won ⎛ Vi Won = nRT ln⎜ ⎜V ⎝ f ⎞ ⎛ Vi ⎟ ⎜ ⇒ ln Q nRT = − ⎟ ⎜V ⎠ ⎝ f ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ The work done on the gas is given by: Substitute for Q in equation (1) to obtain: ⎛ Vi ΔS gas = −nR ln⎜ ⎜V ⎝ f Substitute numerical values and evaluate ΔS: J ⎞ ⎛ 40.0 L ⎞ J ⎛ ⎟ ΔS gas = −(2.00 mol)⎜ 8.314 = 11.5 ⎟ ln⎜ ⎟ ⎜ mol ⋅ K ⎠ ⎝ 80.0 L ⎠ K ⎝ 1826 Chapter 19 (b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings: For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected: The change in entropy of the universe is the change in entropy of the gas: ΔS u = ΔS gas + ΔS surroundings ΔS surroundings = Qrev 0 = =0 T T ΔS u = 11.5 J K 59 •• A 200-kg block of ice at 0.0ºC is placed in a large lake. The temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very slowly. (a) What is the entropy change of the ice? (b) What is the entropy change of the lake? (c) What is the entropy change of the universe (the ice plus the lake)? Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0°C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts. (a) The entropy change of the ice is given by: Substitute numerical values and evaluate ΔSice : ΔS ice = mLf T kJ ⎞ ⎜ (200 kg )⎛ ⎜ 333.5 ⎟ ⎟ ⎝ 273 K kg ⎠ ΔS ice = = 244 kJ K (b) Relate the entropy change of the lake to the entropy change of the ice: (c) The entropy change of the universe due to this melting process is the sum of the entropy changes of the ice and the lake: ΔS lake ≈ −ΔS ice = − 244 kJ K ΔS u = ΔS ice + ΔS lake The Second Law of Thermodynamics 1827 Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is just slightly greater than zero. The melting of the ice is an irreversible process and ΔS u > 0 . 60 •• A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final temperature of the water once thermal equilibrium is established? (b) Find the entropy change of the universe for this process. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe (the entropy change of the piece of ice plus the entropy change of the water in the insulated container). (a) Apply conservation of energy to obtain: ∑Q i ice i =0 or Qmelting + Qwarming − Qcooling = 0 water water Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water: ⎛ kJ ⎞ ⎜ (100 g )⎛ ⎜ 333.5 ⎟ ⎟ + (100 g )⎜ ⎜ 4.18 ⎝ kg ⎠ ⎝ kJ ⎞ ⎟t kg ⋅ C° ⎟ ⎠ ⎛ kJ ⎞ − (100 g )⎜ ⎜ 4.18 kg ⋅ C° ⎟ ⎟ (100°C − t ) = 0 ⎝ ⎠ Solving for t yields: (b) The entropy change of the universe is the sum of the entropy changes of the ice and the water: Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming icewater: t = 10.1°C ΔS u = ΔSice + ΔS water ΔSice = ΔS melting ice + ΔS warming water = ⎛ Tf mLf + mcP ln⎜ ⎜T Tf ⎝ i ⎞ ⎟ ⎟ ⎠ 1828 Chapter 19 Substitute numerical values to obtain: ΔS ice = kJ ⎞ ⎜ (0.100 kg )⎛ ⎟ ⎜ 333.5 ⎟ ⎝ 273 K kg ⎠ ⎛ J kJ ⎞ ⎛ 283 K ⎞ + (0.100 kg )⎜ ⎟ ln⎜ ⎜ 4.18 kg ⋅ K ⎟ ⎟ = 137 K ⎜ 273 K ⎟ ⎠ ⎝ ⎝ ⎠ Find the entropy change of the cooling water: ⎛ J kJ ⎞ ⎛ 283 K ⎞ ΔS water = (0.100 kg )⎜ ⎜ 4.18 kg ⋅ K ⎟ ⎟ ln⎜ ⎜ 373 K ⎟ ⎟ = −115 K ⎝ ⎠ ⎝ ⎠ Substitute for ΔSice and ΔSwater and evaluate the entropy change of the universe: ΔS u = 137 J J J − 115 = 22 K K K Remarks: The result that ΔSu > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe. (a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block: Apply conservation of energy to obtain: ⎛ Tf ΔS Cu = mCu cCu ln⎜ ⎜T ⎝ i ⎞ ⎟ ⎟ ⎠ (1) ∑Q i i =0 or Qcopper + Qwarming = 0 block water 26 K ⎞ ΔS water = (4. and the universe. find the entropy change of the universe.26 K ⎞ ΔS Cu = (1. specific heats.The Second Law of Thermodynamics 1829 Substitute to relate the masses of the block and water to their temperatures. and the final temperature Tf of the water: ⎜ (1.00 kg )⎜ ⎜ 0.00 kg )⎛ ⎜ 0.386 kg ⋅ K ⎟ ⎟ln⎜ ⎜ 373 K ⎟ ⎟ = − 117 K ⎝ ⎠ ⎝ ⎠ (b) The entropy change of the water is given by: ⎛ Tf ΔS water = mwater c water ln⎜ ⎜T ⎝ i ⎞ ⎟ ⎟ ⎠ Substitute numerical values and evaluate ΔS water : ⎛ J kJ ⎞ ⎛ 275. Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead.00 L )⎜1. the water in the lake. 62 •• If a 2.00 kg )⎜ ⎜ 4.26 K Substitute numerical values in equation (1) and evaluate ΔS Cu : ⎛ J kJ ⎞ ⎛ 275. . We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead.386 ⎝ kJ ⎞ ⎟ (Tf − 373 K ) kg ⋅ K ⎟ ⎠ kg ⎞ ⎛ kJ ⎞ ⎛ ⎟ (Tf − 273 K ) = 0 + (4.00 ⎟ ⎜ 4.18 kg ⋅ K ⎟ ⎟ ln⎜ ⎜ 273 K ⎟ ⎟ = 138 K ⎝ ⎠ ⎝ ⎠ (c) Substitute for ΔS Cu and ΔS water and evaluate the entropy change of the universe: ΔS u = ΔS Cu + ΔS water = −117 = 21 J K J J + 138 K K Remarks: The result that ΔSu > 0 tells us that this process is irreversible.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC.18 ⎜ L ⎠⎝ kg ⋅ K ⎟ ⎝ ⎠ Solve for Tf to obtain: Tf = 275. the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. express and evaluate the entropy change of the lead: ⎛ Tf ΔS Pb = mPb cPb ln⎜ ⎜T ⎝ i ⎞ ⎛ J kJ ⎞ ⎛ 283 K ⎞ ⎟ ⎜ 0.41 J/K Substitute numerical values in equation (1) and evaluate ΔSu: ΔS u = −70. (a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs: ΔS u = ΔS h + ΔSc = − ⎛1 1⎞ = −Q⎜ ⎟ ⎜T − T ⎟ c ⎠ ⎝ h Q Q + Th Tc .128 kJ ⎞ ⎟ (90 K ) kg ⋅ K ⎟ ⎠ ⎝ 283 K = 81. (a) What is the change in entropy of the universe.and low-temperature reservoirs.41 = 11 K K K Entropy and ″Lost″ Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K.69 K ⎟ = (2.00 kg )⎜ ⎝ ⎠ ⎝ ⎠ ⎠ The entropy change of the water in the lake is given by: Substitute numerical values and evaluate ΔSw: ΔS w = Qw QPb mPb cPb ΔTPb = = Tw Tw Tw ΔS w = (2.69 J J J + 81.1830 Chapter 19 Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake: ΔS u = ΔS Pb + ΔS w (1) Using the equation for the entropy change during a constant-pressure process. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs. and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high.00 kg )⎛ ⎜ ⎜ 0.128 kg ⋅ K ⎟ ⎟ ln⎜ ⎜ 373 K ⎟ ⎟ = −70. . Picture the Problem Although no energy is lost by the gas in the adiabatic free expansion. the entropy change of the universe is zero. However.3 L to V2 = 24.The Second Law of Thermodynamics 1831 Substitute numerical values and evaluate ΔSu: ⎛ 1 1 ⎞ ΔS u = (− 500 J ) ⎜ ⎜ 400 K − 300 K ⎟ ⎟ ⎝ ⎠ = 0. In the isothermal reversible process that returns the gas to its original state.42 J/K (b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs: The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures: Substitute for εmax to obtain: W = ε max Qh ε max = ε C = 1 − Tc Th ⎛ Tc ⎞ W =⎜ ⎜1 − T ⎟ ⎟Qh h ⎠ ⎝ ⎛ 300 K ⎞ W =⎜ ⎜1 − 400 K ⎟ ⎟ (500 J ) = 125 J ⎝ ⎠ Substitute numerical values and evaluate W: 64 •• In this problem.00 mol of an ideal gas at 300 K undergoes a free adiabatic expansion from V1 = 12. 1. Consequently. (a) What is the entropy change of the universe for the complete cycle? (b) How much work is lost in this cycle? (c) Show that the work lost is TΔSu. the gas releases energy to the surroundings. because the process is reversible. the net entropy change is the negative of that of the gas in the isothermal compression. It is then compressed isothermally and reversibly back to its original state.6 L. the process is irreversible and the entropy of the gas (and the universe) increases. 73 kJ ⎛ Vi ⎞ Wby gas. It operates at 10.763 = 5. isothermal compression free expansion Q T ⎞ ⎟ ⎟ ⎠ The work done by the gas during its isothermal compression is given by: Substituting for Q in the expression for ΔS u and simplifying yields: ⎛ Vf Wby = −Won = −Q = −nRT ln⎜ ⎜V ⎝ i ⎛ Vf ΔS u = − nR ln⎜ ⎜V ⎝ i ⎞ ⎟ ⎟ ⎠ (1) Substitute numerical values and evaluate ΔSu: J J J ⎞ ⎛ 12.0 cycles/s. Application of the first law of thermodynamics will yield the heat given off each cycle.6 L ⎠ ⎝ (b) Use Equation 19-22 to find the amount of energy that becomes unavailable for doing work during this process: (c) No work is done in the free expansion.314 ⎟ln⎜ ⎜ ⎟ K K mol ⋅ K ⎠ ⎝ 24.1832 Chapter 19 (a) Relate the entropy change of the universe to the entropy changes of the gas during 1 complete cycle: ΔS u = ΔS gas during + ΔS gas during free expansion isothermal compresion or. the work done on the gas is: J⎞ ⎛ Wlost = TΔS u = (300 K )⎜ 5.76 ΔS u = −(1.763 ⎟ K⎠ ⎝ = 1. f →i = −Won gas.3 L ⎞ ⎛ ⎟ = 5. (a) How much work is done by the engine during each cycle? (b) How much heat is absorbed from the hot reservoir and how much is released to the cold reservoir during each cycle? Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. In the adiabatic compression. . f →i = −nRT ln⎜ ⎟ ⎜V ⎟ ⎝ f⎠ ⎛ ⎛ Vf ⎞ ⎞ ⎛ Vf ⎞ ⎟ ⎜ = nRT ln⎜ ⎟⎟ ⎜V ⎟ ⎟ = T ⎜ nR ln⎜ ⎜V ⎟ i ⎠ ⎝ ⎝ i⎠ ⎠ ⎝ = TΔS u General Problems 65 • A heat engine with an output of 200 W has an efficiency of 30%.00 mol)⎜ 8. because ΔSgas during ΔS u = ΔS gas during = = 0. 0 J = 67 J 0.The Second Law of Thermodynamics 1833 (a) Use the definition of power to relate the work done in each cycle to the frequency of each cycle: Substitute numerical values and evaluate Wcycle: (b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine: Apply the 1st law of thermodynamics to find the heat given off in each cycle: Wcycle = PΔt = P f where f is the frequency of the engine.cycle = Qh.777 ε C 21.45% 67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K.) Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs.67% = = 0. Wcycle = 200 W = 20.30 Qh.7% 150 J Tc 293 K = 1− = 21.cycle = ε Qc. (a) What is the . a heat engine operating between two heat reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the reservoir at 20ºC. (a) The efficiency of the engine is given by: Substitute numerical values and evaluate ε: (b) Find the efficiency of a Carnot engine operating between the same reservoirs: Express the ratio of the two efficiencies: ε= Q W Qh − Qc = = 1− c Qh Qh Qh ε = 1− 125 J = 16.0 J 10.0 s −1 Wcycle = 20.45% Th 373 K εC = 1− ε 16. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? (This ratio is called the second law efficiency.67% = 16.cycle − W = 67 J − 20 J = 47 J 66 • During each cycle. 510 = 51% 500 K ⎟ ⎠ W = ε Qh = (0. The energy added to the water in the pool is the change in the gravitational potential energy of the diver during the dive. Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25°C.10 MJ Qc.cycle = Qh. (a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: Substitute numerical values and evaluate ε : (b) Use the definition of efficiency to find the work done in each cycle: (c) Apply the first law of thermodynamics to the cycle to obtain: ε = 0.510)(200 kJ ) = 102 kJ = 0.cycle − W = 200 kJ − 102 kJ = 98 kJ 68 • Estimate the change in entropy of the universe associated with an Olympic diver diving into the water from the 10-m platform.85⎜ ⎜1 − ⎝ ⎛ 200 K ⎞ ⎟ = 0.1834 Chapter 19 efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle. The energy added to the water is the change in the gravitational potential energy of the diver: ΔS u = mgh Twater .85ε C = 0. The change in entropy of the universe associated with a dive is given by: Qadded to water Twater where Qadded to water is the energy entering ΔS u = ΔS water = the water as a result of the kinetic energy of the diver as he enters the water.85⎜ ⎜1 − ⎝ ⎛ Tc Th ⎞ ⎟ ⎟ ⎠ ε = 0. the electric power consumption of the electric baseboard heaters is 30. Therefore the entropy of the house does not change: Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house: Substitute for ΔSsurroundings yields: ΔS u = ΔS house + ΔS surroundings or.00 GW of power. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. located on the Hobbes River. and the state of the house does not change. Entropy is a state function. and released by the liquid sodium (and into the superheated steam) in the heat exchanger. ΔS u = ΔS surroundings Q Tsurroundings RΔt Tsurroundings ΔS surroundings = = ΔS u = RΔt Tsurroundings ⇒ ΔS u R = Δt Tsurroundings Substitute numerical values and evaluate ΔSu/Δt: ΔS u 30. generates 1.0 kW on a day when the outside temperature is –7ºC. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. because ΔS house = 0 . (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply . The temperature of the superheated steam is 500 K.81 m/s 2 )(10 m ) ΔS u = (25 + 273)K ≈ 25 J K 69 • To maintain the temperature inside a house at 20ºC.0 kW W = = 113 K Δt 266 K 70 •• Calvin Cliffs Nuclear Power Plant. In this plant. Heat is released into the river. and the water in the river flows by at a temperature of 25ºC. liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine.The Second Law of Thermodynamics 1835 Substitute numerical values and evaluate ΔS u : ( 75 kg ) (9. Heat is absorbed by the liquid sodium in the core. We can find the change in entropy of the surrounding by dividing the heat added by the temperature. 00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river.00 GW 0.00 GW = 1.404 Psupplied = ε max = 2.1836 Chapter 19 1.404 500 K Poutput = 1.4% efficiency. the plant is not allowed to heat the river by more than 0. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt = cΔTρ dV dt . the power that is wasted.00 GW of power and. to produce an output of 1.48 GW dQ dm d = cΔT = cΔT (ρV ) dt dt dt dV = cΔTρ dt dV dQ dt = dt cΔTρ Solve for the flow rate dV/dt of the river: . from this value.48 GW − 1.48 GW Pwasted = Psupplied − Pgenerated (b) Relate the wasted power to the power generated and the power supplied: Substitute numerical values and evaluate Pwasted : (d) Express the rate at which heat is being dumped into the river: Pwasted = 2. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1. at 40.50ºC. (a) The Carnot efficiency of a plant operating between temperatures Tc and Th is given by: Substitute numerical values and evaluate εC: (c) The power that must be supplied.00 GW is given by: ε max = ε C = 1 − Tc Th ε max = 1 − 298 K = 0. Because of these laws. because the efficiency he claims for his invention is greater than the efficiency of a Carnot engine operating between the same two temperatures. He claims that the water vapor absorbs heat at 100°C. What is the minimum rate of exhausting heat that would make you consider believing him? Picture the Problem We can use the inventor’s data to calculate the thermal efficiency of his steam engine and then compare this value to the efficiency of a Carnot engine operating between the same temperatures. (a) Explain to him why he cannot be correct.50 K )⎜10 m 3 ⎟ ⎝ ⎠ ⎝ ⎠ 1. his data is not consistent with what is known about the thermodynamics of engines. you decide he has made an error in the measurement of his exhausted-heat value.The Second Law of Thermodynamics 1837 Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt: J dV s = dt ⎛ J ⎞ ⎛ 3 kg ⎞ ⎜ ⎜ 4180 kg ⎟ ⎟ (0. (a) The Carnot efficiency of an engine operating between these temperatures is: The thermal efficiency of the inventor’s device. in terms of the rate at which it expels heat to the air and does work is: ε C = 1− Tc 298 K = 1− = 20.1% Th 373 K dW dW dt ε = dt = dQh dW dQc + dt dt dt 125 W = = 83. He must have made a mistake in his analysis of his data−or he is a con man looking for suckers to swindle. It is a novel heat engine using water vapor as the working substance.48 × 10 9 = 7. when the air temperature is 25°C.3% 125 W + 25. (b) The maximum efficiency of a steam engine that has ever been achieved is about 50% of the Carnot efficiency of an engine operating between the same temperatures.0 W You should explain to him that. does work at the rate of 125 W.1× 10 5 L/s 71 •• An inventor comes to you to explain his new invention. (b) After careful analysis of the data in his prospectus folder. and releases heat to the air at the rate of only 25.0 W. Setting the efficiency of his steam engine equal to half the Carnot efficiency of the engine yields: dW dW dt 1 ε = dt = 2 C dQh dW dQc + dt dt dt . or (2) A reservoir at 400 K releasing 1. respectively.1838 Chapter 19 Solve for dQc/dt to obtain: ⎞ dW dQc ⎛ 2 =⎜ − 1⎟ ⎜ ⎟ dt ⎝ ε C ⎠ dt Assuming that the inventor has measured the work done per cycle by his invention correctly: dQc ⎛ 2 ⎞ =⎜ − 1⎟ (125 W ) ≈ 1100 W dt ⎝ 0.201)(150 W ) ≈ 15 W Because dQc dQh dW = − . Hint: How much of the 1. What is the efficiency of the cyclic process ABCDA? Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle. let’s assume that his device does take in energy at the rate of 150 W each cycle and find how much work it would do with an efficiency half that of a Carnot engine: Substituting numerical values yields: dW = dt 1 2 (0.0 W of work are done per cycle.00 kJ of heat to a reservoir at 300 K? Explain your choice.0% 750 K 73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K. dW dW 1 dQh 1 ε = dt ⇒ = 2 εC 2 C dQh dt dt dt Ignoring his claim that 125.201 ⎠ a value totally inconsistent with the inventor’s claims for his engine. its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.00 mol of an ideal monatomic gas. The Carnot efficiency of the cycle is given by: Substitute numerical values and evaluate εC: εC = 1− εC = 1− Tc Th 300 K = 60.a dt dt dt reasonable value for dQc/dt is: dQc = 150 W − 15 W = 135 W dt 72 •• The cycle represented in Figure 19-12 (next to Problem 19-14) is for 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? . The temperatures at points A and B are 300 and 750 K. i.833 J/K Helium.00 kJ to find the energy that is lost. is initially at a pressure of 16 atm. (a) For process (2): The efficiency of a Carnot engine operating between temperatures Th and Tc is given by: W2.00 kJ ) = 250 J ⎝ 400 K ⎠ or 750 J are lost. In Part (b) we can use its definition to find the change in entropy for each process. transformed into heat in process (1). Process (1) produces more waste heat.The Second Law of Thermodynamics 1839 Picture the Problem All 500 J of mechanical energy are lost. a volume 74 •• of 1.max = Wrecovered = ε CQin ε C = 1− Tc Th ⎞ ⎟ ⎟Qin ⎠ and hence ⎛ Tc Wrecovered = ⎜ ⎜1 − T h ⎝ Substitute for ε C to obtain: ⎛ 300 K ⎞ Wrecovered = ⎜1 − ⎟(1..0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. a monatomic gas. It is quasi-statically expanded at constant temperature until its volume is 4. (a) Sketch this .00 kJ )⎜ ⎜ 300 K − 400 K ⎟ ⎟ ⎝ ⎠ = 0. (b) Find the change in entropy of the universe for process (1): Express the change in entropy of the universe for process (2): ΔS1 = ΔQ 500 J = = 1.0 L.67 J/K T 300 K ΔQ ΔQ + Th Tc ΔS 2 = ΔS h + ΔSc = − ⎛1 1⎞ = ΔQ ⎜ ⎜T − T ⎟ ⎟ h ⎠ ⎝ c Substitute numerical values and evaluate ΔS2: ⎛ 1 1 ⎞ ΔS 2 = (1.e. and a temperature of 600 K. we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1. For process (2). Process (2) is more wasteful of available work. 2. To find the work done during each cycle.67) to relate the temperatures and volumes at 1 and 3: T3V3 γ −1 = T1V1 γ −1 ⎛ V1 ⎞ ⇒ T3 = T1 ⎜ ⎜V ⎟ ⎟ ⎝ 3⎠ γ −1 . (c) Find the work done during each step of the cycle. volumes. isobaric.67 = 2.0 atm V2 ⎝ ⎠ 1γ ⎛P ⎞ ⎜ 1⎟ P 1V1 = P 3V3 ⇒ V3 = V1 ⎜ ⎟ ⎝ P3 ⎠ γ γ ⎛ 16 atm ⎞ V3 = (1. (d) Find the efficiency of the cycle. and adiabatic processes.294 L Apply an equation for an adiabatic process (γ =1. (b) Find the volume and temperature after the compression at constant pressure.1840 Chapter 19 cycle on a PV diagram.0 L ⎟ ⎟ = 4. (a) The PV diagram of the cycle is shown to the right. we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion.3 L 1 1. and 3 with 1 being the initial state. we can use the equations for the work done during isothermal. and 3. and pressures at points 1. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures.0 atm ⎟ ⎟ ⎝ ⎠ = 2.0 L ⎞ V1 = (16 atm )⎜ ⎜ 4. Finally. (b) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2: Apply an equation for an adiabatic process to relate the pressures and volumes at 1 and 3: Substitute numerical values and evaluate V3: P2 = P1 ⎛ 1. 2. Picture the Problem Denote the three states of the gas as 1.0 L )⎜ ⎜ 4. 294 L − 4. The volume of the balloon increases by 4. where Tc < 120ºC.00 L.116 atm ⋅ L = 5 atm ⋅ L (d) Use its definition to express the efficiency of the cycle: Substitute numerical values and evaluate ε: ε= W W W = = Qin Q12 W12 5.0 L )ln⎜ ⎜ 1.18 atm ⋅ L ε= 75 •• [SSM] A heat engine that does the work of blowing up a balloon at a pressure of 1.294 L )] 2 [( = −10.0 atm )(2.24 atm ⋅ L Substitute numerical values in equation (1) and evaluate W: W = 22. and heat is released to a reservoir at a temperature Tc.24 atm ⋅ L = 5.00 L ) = −6.0 L ⎞ T3 = (600 K )⎜ ⎜ 2.294 L ⎟ ⎟ ⎝ ⎠ = 3.The Second Law of Thermodynamics 1841 Substitute numerical values and evaluate T3: ⎛ 1. . find the temperature Tc.0 L ) − (4.824 atm ⋅ L For the process 3→1: 3 W31 = −C V ΔT31 = − 3 1V1 − P 3V3 ) 2 nR (T1 − T3 ) = − 2 (P =−3 16 atm )(1.00 kJ from a reservoir at 120ºC.116 atm ⋅ L ≈ 20% 22.0 atm )(2.18 atm ⋅ L For the process 2→3: W23 = P2 ΔV23 = (4.0 L ⎟ ⎟ ⎝ ⎠ = 22.824 atm ⋅ L − 10.18 atm ⋅ L − 6.4 ×10 2 K (c) Express the work done each cycle: For the process 1→2: W = W12 + W23 + W31 (1) 1. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs.00 atm absorbs 4.0 L ⎞ = (16 atm )(1.67 −1 = 344 K ⎛ V2 ⎞ ⎛ V2 ⎞ ⎟ ⎜ W12 = nRT1 ln⎜ = P V ln 1 1 ⎜V ⎟ ⎜V ⎟ ⎟ ⎝ 1⎠ ⎝ 1⎠ ⎛ 4. relate the heat removed from the cold reservoir to the work done each cycle: COP = Qc W .00 atm )(4. relate Tc to the work done by and the heat input to the real heat engine.1842 Chapter 19 Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and.00 atm ⋅ L Substitute numerical values and evaluate Tc: ⎛ 101. given that the efficiency of the heat engine is half that of a Carnot engine. relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir: Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures: Substitute for ε C to obtain: εC = 1− Tc ⇒ Tc = Th (1 − ε C ) Th ε= 2W W =1 2 εC ⇒ εC = Qin Qin ⎛ 2W Tc = Th ⎜ ⎜1 − Q in ⎝ ⎞ ⎟ ⎟ ⎠ The work done by the gas in expanding the balloon is: W = PΔV = (1.00 atm ⋅ L × ⎟⎟ atm ⋅ L ⎠ ⎟ ⎝ ⎜ Tc = (393 K ) 1 − = 313 K ⎜ ⎟ 4. Picture the Problem We can use the definitions of the COP and εC to show that their relationship is ε C × COPC = TC Th . Using the definition of the COP. Using its definition.00 kJ ⎜ ⎟ ⎝ ⎠ Show that the coefficient of performance of a Carnot engine run as a 76 •• refrigerator is related to the efficiency of a Carnot engine operating between the same two temperatures by ε C × COPC = Tc Th .00 L ) = 4.325 J ⎞ ⎞ ⎛ ⎜ 2 ⎜ 4. and W: Substitute for Qc to obtain: Qc = Qh − W Qh − W W 1− COP = Divide the numerator and denominator by Qh and simplify to obtain: W Q −W Qh COP = h = W W Qh ⎛ Tc 1− ⎜ ⎜1 − T 1− ε C h = ⎝ COPc = ⎞ Tc ⎟ ⎟ T ⎠= h Because ε C = T W = 1− c : Qh Th εC εC εC and ε C × COPc = Tc Th 77 •• A freezer has a temperature Tc = –23ºC. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer. relate the heat that must be removed from the freezer to the work done by the motor: Differentiate this expression with respect to time to express the power of the motor: Express the maximum COP of the motor: COP = Q Qc ⇒W = c W COP P= dW dQc dt = dt COP COPmax = Tc ΔT . Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. The air in the kitchen has a temperature Th = 27ºC. Qh. The freezer is not perfectly insulated and some heat leaks through the walls of the freezer at a rate of 50 W. Find the power of the motor that is needed to maintain the temperature in the freezer. Using the definition of the COP.The Second Law of Thermodynamics 1843 Apply energy conservation to relate Qc. 325 kPa 5. pressures. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression.00 mol)⎛ ⎜ 8.314 TB = TA VB 2V = (600 K ) A VA VA = 1200 K . (a) Apply the ideal-gas law to find the volume of the gas at A: VA = nRTA PA J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ = 101. The volume at B is twice the volume at A.4 L (2. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures.) At A the pressure and temperature are 5. and volumes at points A.969 ×10 − 2 m 3 × −3 3 = 19.38 L = 39.69 L 10 m = 19.00 atm and 600 K. adiabatic.7 L (b) We’re given that VB = 2VA . and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constantpressure.69 L ) = 39. 2. B. Hence: Apply the ideal-gas law to this constant-pressure process to obtain: VB = 2(19. (The PV diagram is not drawn to scale.00 atm × atm 1L = 1.1844 Chapter 19 Substitute for COPmax to obtain: P= dQc ΔT dt Tc Substitute numerical values and evaluate P: ⎛ 50 K ⎞ P = (50 W )⎜ ⎜ 250 K ⎟ ⎟ = 10 W ⎝ ⎠ 78 •• In a heat engine. and isothermal processes of the cycle.00 mol of a diatomic gas are taken through the cycle ABCA as shown in Figure 19-20. 69 L ) = 98.69 L ⎞ J ⎞ ⎛ ⎟ ⎜ ( ) ( ) = 2.77 L ⎟ ⎟ ⎟ mol ⋅ K ⎠ ⎝ ⎝ ⎠ ⎠ = −24.494 ×10 J 2 mol ⋅ K ⎠ ⎝ = 24.314 600 K ln ⎜ ⎟ ⎜ 222. BC − 0 = ΔEint. BC = ΔEint.9754 ×10 3 J = 9. BC = −ncV ΔTBC =−5 nRΔTBC 2 Substitute numerical values and evaluate WBC: J ⎞ 4 (2.20 kJ = − 24.2 kJ . BC − Qin.77 L ⎝ ⎠ = 223 L 1 (e) The work done by the gas during the constant-pressure process AB is given by: Substitute numerical values and evaluate WAB: WAB = PA (VB − VA ) = PA (2VA − VA ) = PAVA WAB = (5.45 atm ⋅ L × 101.314 ⎟ (600 K − 1200 K ) = 2.00 atm )(19.00 mol)⎛ WBC = − 5 ⎜ 8.38 L )⎜ ⎜ 600 K ⎟ ⎟ = 222.00 mol 8.4 −1 VC = (39.4) to find the volume of the gas at C: Substitute numerical values and evaluate VC: TC = TA = 600 K 1 ⎛ TB ⎞ γ −1 γ −1 TBVB = TCVCγ −1 ⇒ VC = VB ⎜ ⎜T ⎟ ⎟ ⎝ C⎠ ⎛ 1200 K ⎞ 1.325 J atm ⋅ L = 9.98 kJ Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC: WBC = ΔEint.The Second Law of Thermodynamics 1845 (c) Because the process C→A is isothermal: (d) Apply an equation for an adiabatic process (γ = 1.9 kJ The work done by the gas during the isothermal compression CA is: ⎛ VA WCA = nRTC ln⎜ ⎜V ⎝ C ⎞ ⎛ 19. and volumes at points B.) The segment AB represents an isothermal expansion.00 atm and 600 K.50 atm × = 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C: TC = TB PCVC PBVB (1) . CA = 0 for an isothermal process. and D.00 atm.00 atm ) A VB 2VA 101. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the pressure at B: PB = PA VA V = (5.9 kJ The heat absorbed during the adiabatic expansion BC is: 7 2 (2. the segment BC an adiabatic expansion. (The PV diagram is not drawn to scale.2 kJ because ΔEint.00 mol)⎛ ⎜ 8. C. The pressure at D is 1.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. 79 •• [SSM] In a heat engine.314 ⎝ J ⎞ ⎟ (1200 K − 600 K ) mol ⋅ K ⎠ QBC = 0 Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA: QCA = WCA + ΔEint. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. The volume at B is twice the volume at A.3 kPa 1atm = 2.92 kJ = 34. CA = WCA = − 24. 2.325 kPa = 253.1846 Chapter 19 (f) The heat absorbed during the constant-pressure expansion AB is: QAB = ncP ΔTA − B = 7 nRΔTA − B = 2 = 34. Picture the Problem We can use the ideal-gas law to find the unknown temperatures. The pressure and temperature at A are 5. pressures. 39 L )⎜ ⎜ 1.09 atm ⋅ L × = −5.39 L ) (c) The work done by the gas in one cycle is given by: W = WAB + WBC + WCD + WDA The work done during the isothermal expansion AB is: ⎛ VB WAB = nRTA ln⎜ ⎜V ⎝ A ⎞ ⎛ 2VA J ⎞ ⎛ ⎟ ⎟ = (2.4 TC = (600 K ) = 462 K (1.00 mol)⎛ WBC = −C V ΔTBC = − 5 nRΔTBC = − 5 ⎜ 8.39 L Use the equation of state for an adiabatic process and γ = 1.00 atm ⎟ ⎟ ⎝ ⎠ 1 1.680 kJ 101.314 mol ⋅ K ⎟ (600 K ) ln⎜ ⎜ V ⎝ ⎠ ⎠ ⎝ A ⎞ ⎟ ⎟ = 6.3 kPa ⎛ 2.4 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate TC: ⎛ PB ⎞ VC = VB ⎜ ⎜P ⎟ ⎟ ⎝ C⎠ = 75.00 mol)⎛ ⎜ 8.314 = = 39.325 J atm ⋅ L Express and evaluate the work done during the constant-volume process DA: WDA = 0 .78 L 1γ J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 253.00 atm )(19.00 atm )(75.50 atm )(39.737 kJ The work done during the isobaric compression CD is: WCD = PC (VD − VC ) = (1.The Second Law of Thermodynamics 1847 Use the ideal-gas law to find the volume of the gas at B: VB = nRTB PB (2.78 L) (2.00 mol)⎜ 8.78 L ) = −56.314 ⎟ (462 K − 600 K ) 2 2 mol ⋅ K ⎠ ⎝ = 5.915 kJ ⎠ The work done during the adiabatic expansion BC is: J ⎞ (2.50 atm ⎞ = (39.7 L − 75. 7 L (b) We’re given that: VB = 2VA = 2(19. adiabatic.) At A the pressure and temperature are 5.39 L = 39.680 kJ + 0 = 6. and isothermal processes of the cycle. B. The volume at B is twice the volume at A. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed by the gas in each segment of the cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures. and volumes at points A.4 L Apply the ideal-gas law to this isobaric process to find the temperature at B: (c) Because the process CA is isothermal: (d) Apply an equation for an adiabatic process (γ = 5/3) to express the volume of the gas at C: TB = TA VB 2V = (600 K ) A VA VA = 1200 K TC = TA = 600 K ⎛ TB ⎞ γ −1 ⇒ VC = VB ⎜ ⎜T ⎟ ⎟ ⎝ C⎠ 1 TBVB γ −1 = TCVC γ −1 .69 L = 19.97 kJ 80 •• In a heat engine.325 kPa 5. (a) Apply the ideal-gas law to find the volume of the gas at A: VA = nRTA PA (2. 2. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression.314 = J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 101.737 kJ − 5.00 atm × atm = 19.00 mol)⎛ ⎜ 8.00 atm and 600 K.00 mol of a monatomic gas are taken through the cycle ABCA as shown in Figure 19-20. pressures.915 kJ + 5.1848 Chapter 19 Substitute numerical values and evaluate W: W = 6.69 L ) = 39.972 kJ = 6. (The PV diagram is not drawn to scale. and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric. BC = ΔEint.00 mol 8.4 L = 111 L (e) The work done by the gas during the isobaric process AB is given by: Substitute numerical values and evaluate WAB: WAB = PA (VB − VA ) = PA (2VA − VA ) = PAVA WAB = (5.975 kJ = 9.97 kJ 2 mol ⋅ K ⎠ ⎝ = 15.00 mol)⎛ ⎜ 8.314 ⎟ (600 K − 1200 K ) = 14.9 kJ 5 2 (2.29 kJ − 17.0 kJ The work done by the gas during the isothermal compression CA is: ⎛ VA WCA = nRTC ln⎜ ⎜V ⎝ C ⎞ ⎛ 19.98 kJ Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC: WBC = ΔEint.00 mol)⎛ WBC = − 3 ⎜ 8.39 L )⎜ ⎜ 600 K ⎟ ⎟ ⎝ ⎠ = 111.314 600 K ln ⎜ ⎟ ⎜ 111. BC − 0 =−3 nRΔTBC 2 Substitute numerical values and evaluate WBC: J ⎞ (2.69 L ) = 98.69 L ⎞ J ⎞ ⎛ ⎟ ⎜ ( ) ( ) = 2. AB = ncP ΔTAB = 5 nRΔTAB = 2 = 24.45 atm ⋅ L × 101.325 J atm ⋅ L 3 = 9. BC − Qin. BC = −(ncV ΔTBC ) = −17.314 ⎝ J ⎞ ⎟ (1200 K − 600 K ) mol ⋅ K ⎠ .00 atm )(19.The Second Law of Thermodynamics 1849 Substitute numerical values and evaluate VC: ⎛ 1200 K ⎞ 2 VC = (39.4 L ⎟ ⎟ ⎟ mol ⋅ K ⎠ ⎝ ⎝ ⎠ ⎠ = ΔEint.3 kJ (f) The heat absorbed during the isobaric expansion AB is: Qin. pressures. CA = WCA = − 17. (The PV diagram is not drawn to scale. Picture the Problem We can use the ideal-gas law to find the unknown temperatures. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB: PB = PA VA V = (5. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle.325 kPa 1atm = 2. the segment BC an adiabatic expansion.3 kJ because ΔEint = 0 for an isothermal process. 81 •• In a heat engine. The pressure and temperature at A are 5. 2.3 kPa = 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC: Use the ideal-gas law to find the volume at B: TC = TB PCVC PBVB (1) VB = nRTB PB (2.00 atm and 600 K.00 atm ) A 2VA VB 101.3 kPa . and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.39 L J ⎞ ⎟ (600 K ) mol ⋅ K ⎠ ⎝ 253.314 = = 39. C. and volumes at points B.00 mol)⎛ ⎜ 8.00 atm.00 mol of a monatomic gas are carried through the cycle ABCDA shown in Figure 19-21. The volume at B is twice the volume at A.50 atm × = 253. The pressure at D is 1.1850 Chapter 19 Express and evaluate the heat absorbed during the adiabatic expansion BC: Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA: QBC = 0 QCA = WCA + ΔEint.) The segment AB represents an isothermal expansion. ) .00 mol)⎜ 8.00 atm ⎟ ⎟ ⎝ ⎠ 35 (1.56 atm ⋅ L × = −4.9 K = 416 K (c) The work done by the gas in one cycle is given by: W = WAB + WBC + WCD + WDA (2) The work done during the isothermal expansion AB is: ⎛ VB WAB = nRTA ln⎜ ⎜V ⎝ A ⎞ ⎛ 2VA J ⎞ ⎛ ⎟ ⎟ = (2. (The Otto cycle is discussed in Section 19-1.39 L ) = 415.00 atm )(68.7 L − 68.920 kJ 101.The Second Law of Thermodynamics 1851 Use the equation of state for an adiabatic process and γ = 5/3 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate TC: ⎛ PB ⎞ VC = VB ⎜ ⎜P ⎟ ⎟ ⎝ C⎠ = 68.592 kJ − 4.59 kJ 82 •• Compare the efficiency of the Otto cycle to the efficiency of the Carnot cycle operating between the same maximum and minimum temperatures.39 L )⎜ ⎜ 1.592 kJ The work done during the isobaric compression CD is: WCD = PC (VD − VC ) = (1.00 atm )(19.314 mol ⋅ K ⎟ (600 K ) ln⎜ ⎜V ⎝ ⎠ ⎠ ⎝ A ⎞ ⎟ ⎟ = 6.9 K − 600 K ) 2 mol ⋅ K ⎠ ⎝ = 4.00 mol)⎛ =−3 ⎜ 8.26 L ) (2.915 kJ ⎠ The work done during the adiabatic expansion BC is: WBC = −C V ΔTBC = − 3 nRΔTBC 2 J ⎞ (2.50 atm )(39.915 kJ + 4.314 ⎟ (415.920 kJ + 0 = 6.325 J atm ⋅ L The work done during the constantvolume process DA is: Substitute numerical values in equation (2) to obtain: WDA = 0 W = 6.26 L ) = −48.26 L TC = (600 K ) 1γ ⎛ 2.50 atm ⎞ = (39. while Ta is the lowest temperature of the cycle. Substitute to obtain: ⎛ Va ⎞ Tb = Ta ⎜ ⎜V ⎟ ⎟ ⎝ b⎠ ⎛ Vd Tc = Td ⎜ ⎜V ⎝ c γ −1 ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Vd Tc − Tb = Td ⎜ ⎜V ⎝ c ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Va ⎞ − Ta ⎜ ⎜V ⎟ ⎟ ⎝ b⎠ γ −1 ⎛ Va ⎞ Tc − Tb = Td ⎜ ⎜V ⎟ ⎟ ⎝ b⎠ γ −1 ⎛ Va ⎞ − Ta ⎜ ⎜V ⎟ ⎟ ⎝ b⎠ γ −1 γ −1 ⎛ Va ⎞ = (Td − Ta )⎜ ⎜V ⎟ ⎟ ⎝ b⎠ Substitute in equation (1) and simplify to obtain: ε Otto = 1 − Td − Ta ⎛ Va (Td − Ta )⎜ ⎜V ⎝ b γ −1 ⎞ ⎟ ⎟ ⎠ γ −1 ⎛ Vb ⎞ Ta = 1− ⎜ ⎟ = 1− T ⎜V ⎟ b ⎝ a⎠ Note that. We can apply the relation TV γ −1 = constant to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d.1852 Chapter 19 Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. Va = Vd and Vc = Vb. Tb is not the highest temperature. . We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. The efficiency of the Otto engine is given in Example 19-2: ε Otto = 1 − Td − Ta Tc − Tb (1) where the subscripts refer to the various points of the cycle as shown in Figure 19-3. Apply the relation TV γ −1 = constant to the adiabatic process a→b to obtain: Apply the relation TV γ −1 = constant to the adiabatic process c→d to obtain: Subtract the first of these equations from the second to obtain: In the Otto cycle. we can compare the efficiencies by examining their ratio. Finally. is the Brayton cycle. often used in refrigeration.The Second Law of Thermodynamics 1853 Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc: The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by: Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures: P Pc Pb = ⇒ Tc = Tb c > Tb Pb Tc Tb ε Carnot = 1 − Ta Tc ε Carnot ε Otto Ta Tc > 1 because Tc > Tb. which involves (1) an adiabatic compression. where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. P 2 Qh ⇓ 3 1 ⇓ 4 V Qc . Assume the system begins the adiabatic compression at temperature T1. and transitions to temperatures T2.(3) an adiabatic expansion. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. and (4) an isobaric compression back to the original state. (a) Sketch this cycle on a PV diagram. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. ε Carnot > ε Otto 83 ••• [SSM] A common practical cycle. (c) Show that this (T3 − T2 ) efficiency. Because the adiabatic transitions in the cycle do not have heat flow associated with them. (a) The Brayton heat engine cycle is shown to the right. (2) an isobaric (constant pressure) expansion. can be written as ε = 1 − r (1− γ ) γ . all we must do is consider the heat flow in and out of the engine during the isobaric transitions. = Ta 1− Tb 1− Hence. T3 and T4 after each leg of the cycle. (b) Show that the (T − T ) efficiency of the overall cycle is given by ε = 1 − 4 1 . The paths 1→2 and 3→4 are adiabatic. for an adiabatic transition. use the ideal-gas law to eliminate V and obtain: Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2: Similarly.1854 Chapter 19 (b) The efficiency of a heat engine is given by: During the constant-pressure expansion from state 1 to state 2 heat enters the system: During the constant-pressure compression from state 3 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields: ε= Q − Qc W = h Qin Qin (1) Q23 = Qh = nC P ΔT = nC P (T3 − T2 ) Q41 = −Qc = −nC P ΔT = −nC P (T1 − T4 ) ε= = (T3 − T2 ) + (T1 − T4 ) (T3 − T2 ) (T − T ) = 1− 4 1 (T3 − T2 ) (c) Given that. for the adiabatic transition from state 3 to state 4: nC P (T3 − T2 ) − (− nC P (T1 − T4 )) nC P (T3 − T2 ) Tγ = constant P γ −1 ⎛P T1γ T2γ ⇒ T1 = ⎜ low = γ −1 γ −1 ⎜P Plow Phigh ⎝ high ⎞ ⎟ ⎟ ⎠ γ −1 γ T2 ⎛P T4 = ⎜ low ⎜P ⎝ high ⎞ ⎟ ⎟ ⎠ γ −1 γ T3 γ −1 γ γ −1 γ Subtract T1 from T4 and simplify to obtain: ⎛P T4 − T1 = ⎜ low ⎜P ⎝ high ⎛P = ⎜ low ⎜P ⎝ high ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎛P T3 − ⎜ low ⎜P ⎝ high ⎞ ⎟ ⎟ ⎠ T2 γ −1 γ (T3 − T2 ) . TV γ −1 = constant . until its temperature is T3.0 h each day. The pressure ratio r = Phigh/Plow for the cycle is 5.The Second Law of Thermodynamics 1855 Dividing both sides of the equation by T3 − T2 yields: P T4 − T1 ⎛ = ⎜ low ⎜ T3 − T2 ⎝ Phigh ⎞ ⎟ ⎟ ⎠ γ −1 γ Substitute in the result of Part (b) and simplify to obtain: ⎛P ε = 1 − ⎜ low ⎜P ⎝ high = 1 − (r ) γ where r = ⎞ ⎟ ⎟ ⎠ γ −1 γ ⎛ Phigh = 1− ⎜ ⎜P ⎝ low ⎞ ⎟ ⎟ ⎠ 1−γ γ 1−γ Phigh Plow 84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. what is the rate at which electrical energy must be supplied to the motor of this refrigerator? (e) Assuming the refrigerator motor is actually running for only 4. Then the gas is adiabatically compressed.0. Finally. The cylinder containing the refrigerant (a monatomic gas) has an initial volume and pressure of 60 mL and 1. In this case. how much does it add to your monthly electric bill. (T4 − T1 ) . .0 atm. the volume and temperature are 75 mL and –25°C. Assume 15 cents per kWh of electric energy and thirty days in a month. And then it is compressed at constant pressure until its temperature T2. all we must do is consider the heat flow in and out of the refrigerator during the isobaric transitions. Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. (a) Sketch this cycle on a PV diagram. the cycle begins at temperature T1 and expands at constant pressure until its temperature T4. (b) Show that the coefficient of performance is COPB = (T3 − T2 − T4 + T1 ) (c) Suppose your ″Brayton cycle refrigerator″ is run as follows. After the expansion at constant pressure. Because the adiabatic transitions in the cycle do not have heat flow associated with them. What is the coefficient of performance for your refrigerator? (d) To absorb heat from the food compartment at the rate of 120 W. it adiabatically expands until it returns to its initial state at temperature T1. Heat Qc enters the gas during the constantpressure transition from state 1 to state 4 and heat Qh leaves the gas during the constant-pressure transition from state 3 to state 2. The paths 1→2 and 3→4 are adiabatic. P Qh 3 4 V (1) Qc . and 4. the quotient T/V is constant: Substitute numerical values and evaluate T1: T4 = −25°C + 273 K = 248 K ⎛ V1 ⎞ T1 T4 = ⇒ T1 = ⎜ ⎜V ⎟ ⎟T4 V1 V4 ⎝ 4⎠ ⎛ 60 mL ⎞ T1 = ⎜ ⎟ (248 K ) = 198 K ⎝ 75 mL ⎠ ⇓ ⇓ (a) The Brayton refrigerator cycle is shown to the right. 2. We’re given that the temperature in state 4 is: For the constant-pressure transition from state 1 to state 4.1856 Chapter 19 2 1 Qc W where W = Q32 − Q14 COPB = (b) The coefficient of performance of the Brayton cycle refrigerator is given by: During the constant-pressure compression from state 3 to state 2 heat leaves the system: During the constant-pressure expansion from state 1 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields: Q32 = −Qh = −nC P ΔT = − nC P (T2 − T3 ) Q14 = Qc = nC P ΔT = nC P (T4 − T1 ) COPB = = = (T4 − T1 ) − (T4 − T1 ) − (T2 − T3 ) T4 − T1 T3 − T2 − T4 + T1 nCP (T4 − T1 ) − nCP (T2 − T3 ) − nCP (T4 − T1 ) (c) The COPB requires the temperatures corresponding to states 1. 3. 67 (248 K ) = 473 K γ −1 1.The Second Law of Thermodynamics 1857 Given that. TV γ −1 = constant . if the frequency of the AC power input is f. use the ideal-gas law to eliminate V and obtain: For the adiabatic transition from state 4 to state 3: Tγ = constant P γ −1 ⎛ P3 ⎞ T3γ T4γ ⇒ T3 = ⎜ = γ −1 γ −1 ⎜P ⎟ ⎟ P3 P4 ⎝ 4⎠ 1. for an adiabatic transition.67 −1 ⎛ P2 ⎞ γ 1.1 W= Qc COPB The rate at which energy must be supplied to this refrigerator is given by: 1 dQc dW = dt COPB dt or.67 ( ⎟ ( ) T2 = ⎜ = T 5 198 K ) 1 ⎜P ⎟ ⎝ 1⎠ = 378 K Substitute numerical values in the expression derived in Part (a) and evaluate COPB: (d) From the definition of COPB: COPB = 248 K − 198 K 473 K − 378 K − 248 K + 198 K = 1. fQc dW = dt COPB Express the heat Qc that is drawn from the cold reservoir: Substituting for Qc yields: Qc = nC P ΔT = nC P (T4 − T1 ) fnC P (T4 − T1 ) dW = COPB dt n= P4V4 RT4 Use the ideal-gas law to express the number of moles of the gas: .67 −1 γ −1 γ T4 Substitute numerical values and evaluate T3: Similarly. for the adiabatic transition from state 2 to state 1: T3 = (5) 1. Picture the Problem We can use nR = CP − CV .21 kW (e) The monthly cost of operation is given by Monthly Cost = Cost Per Unit of Power × Power Consumption = rate × daily consumption × number of days per month Substitute numerical values and evaluate the monthly cost of operation: Monthly Cost = $0.T1) to state (V2. show explicitly that the entropy change is zero for a quasi-static adiabatic expansion from state (V1. Substitute for n and CP to 2 obtain: 5 2 dW = dt = f 5 2 fP4V4 (T4 − T1 ) (COPB )T4 P4V4 R(T4 − T1 ) RT4 COPB Substitute numerical values and evaluate dW/dt: 5 2 dW = dt (60 s ) −1 ⎛ 10 −3 m 3 ⎞ ⎜ ⎟ (101. T1) to state (V2.207 kW × × 30 d ≈ $4 kWh d 85 ••• Using ΔS = CV ln (T2 T1 ) + nR ln (V2 V1 ) (Equation 19-16) for the entropy change of an ideal gas. γ = CP CV .1858 Chapter 19 Because the gas is monatomic.11)(248 K ) = 0.0 h × 0. Express the entropy change for a general process that proceeds from state 1 to state 2: For an adiabatic process: ⎛ T2 ⎞ ⎛ V2 ⎞ ΔS = CV ln⎜ ⎜T ⎟ ⎟ + nR ln⎜ ⎜V ⎟ ⎟ ⎝ 1⎠ ⎝ 1⎠ T2 ⎛ V1 ⎞ =⎜ ⎟ ⎟ T1 ⎜ ⎝ V2 ⎠ γ −1 .T2) is zero. and TV γ −1 = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1. T2). CP = 5 R .15 4.325 kPa )⎜ 75 mL × ⎟(248 K − 198 K ) L ⎝ ⎠ = 207 W (1. then the entropy change of the universe is ΔSu = −Qh/Th + 0. (b) Show that if the heat-engine statement of the second law were not true. heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir. that is. Su < 0. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal amount of heat is transferred to the hot reservoir and W = 0. then the entropy of the universe could decrease. . which is negative. i. then the entropy of the universe could decrease.The Second Law of Thermodynamics 1859 Substitute for T2 and simplify to obtain: T1 γ −1 ⎡ ⎛ V1 ⎞ ⎤ ⎢ CV ln⎜ γ −1 ⎜V ⎟ ⎟ ⎥ ⎛ V1 ⎞ ⎛ V2 ⎞ ⎛ V2 ⎞ ⎢ ⎥ 2 ⎠ ⎝ ΔS = CV ln⎜ ⎜V ⎟ ⎟ + nR ln⎜ ⎜V ⎟ ⎟ = ln⎜ ⎜V ⎟ ⎟ ⎢nR + ⎥ V ⎝ 2⎠ ⎝ 1⎠ ⎝ 1 ⎠⎢ ln 2 ⎥ V1 ⎢ ⎥ ⎣ ⎦ ⎡ ⎛ V1 ⎞ ⎤ (γ − 1)CV ln⎜ ⎢ ⎜V ⎟ ⎟⎥ ⎛ V2 ⎞ ⎛ V2 ⎞ ⎢ 2 ⎠⎥ ⎝ = ln⎜ = ln⎜ ⎜V ⎟ ⎟ [nR − (γ − 1)CV ] ⎜V ⎟ ⎟ ⎢nR + V1 ⎥ 1 ⎝ ⎠ ⎝ 1⎠ − ln ⎢ ⎥ V 2 ⎢ ⎥ ⎣ ⎦ Use the relationship between CP and CV to obtain: Substituting for nR and γ and simplifying yields: nR = CP − CV ⎛ Cp ⎞ ⎤ ⎛ V2 ⎞ ⎡ ⎜ ⎟ ⎟ ΔS = ln⎜ C − C − − 1 ⎢ P V ⎜V ⎟ ⎜C ⎟CV ⎥ ⎝ 1 ⎠⎣ ⎝ V ⎠ ⎦ = 0 86 ••• (a) Show that if the refrigerator statement of the second law of thermodynamics were not true. the entropy of the universe would decrease. Qc = 0. Again. Because Th > Tc. the entropy of the universe would decrease. (b) In this case.. (c) A third statement of the second law is that the entropy of the universe cannot decrease.e. The entropy change of the universe is then ΔSu = Qc/Th − Qc/Tc. respectively. W2. Show that the net efficiency of the combination is given by ε net = ε 1 + ε 2 − ε 1ε 2 . The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ΔSu ≥ 0. such that the heat released by the first engine is used as the heat absorbed by the second engine. where Th > Tm > Tc. Engine 1 operates between temperatures Th and Tm and Engine 2 operates between Tm and Tc. and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1. The efficiencies of the engines are ε1 and ε2. Suppose that each engine is an ideal reversible heat engine. Picture the Problem We can express the net efficiency of the two engines in terms of W1. W2. such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. Show that the net efficiency of T the combination is given by ε net = 1 − c . Express the net efficiency of the two heat engines connected in series: Express the efficiencies of engines 1 and 2: Solve for W1 and W2 and substitute to obtain: Express the efficiency of engine 1 in terms of Qm and Qh: Substitute for Qm/Qh and simplify to obtain: ε net = W1 + W2 Qh ε1 = W1 W and ε 2 = 2 Qh Qm ε net = ε1Qh + ε 2Qm Qh = ε1 + Qm ε2 Qh ε1 = 1 − Qm Q ⇒ m = 1 − ε1 Qh Qh ε net = ε 1 + (1 − ε 1 )ε 2 = ε 1 + ε 2 − ε 1ε 2 88 ••• Suppose that two heat engines are connected in series. as shown in Figure 19-22. Qh. 87 ••• Suppose that two heat engines are connected in series. but these statements do not specify the minimum amount of heat rejected or work that must be done. (Note that this result means that two Th . and Qm. The statement ΔSu ≥ 0 is more restrictive.1860 Chapter 19 (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir. and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1. Finally. Qh. Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii): . and Qm. we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain ε net = 1 − Tc Th . W2. W2.The Second Law of Thermodynamics 1861 reversible heat engines operating ″in series″ are equivalent to one reversible heat engine operating between the hottest and coldest reservoirs.) Picture the Problem We can express the net efficiency of the two engines in terms of W1. Express the efficiencies of ideal reversible engines 1 and 2: ε1 = 1 − and Tm Th Tc Tm (1) ε2 = 1− The net efficiency of the two engines connected in series is given by: Express the efficiencies of engines 1 and 2: Solve for W1 and W2 and substitute in equation (3) to obtain: Express the efficiency of engine 1 in terms of Qm and Qh: Substitute for Qm to obtain: Qh (2) ε net = ε1 = W1 + W2 Qh (3) W1 W and ε 2 = 2 Qh Qm ε net = ε1Qh + ε 2Qm Qh = ε1 + Qm ε2 Qh ε1 = 1 − Q Qm ⇒ m = 1 − ε1 Qh Qh ε net = ε1 + (1 − ε1 )ε 2 Substitute for ε1 and ε2 and simplify to obtain: ε net = 1 − Tm ⎛ Tm ⎞ ⎛ Tc ⎞ +⎜ ⎟ ⎟⎜ ⎜1 − T ⎟ ⎟ Th ⎜ T h m ⎠ ⎝ ⎠⎝ T T T T = 1− m + m − c = 1− c Th Th Th Th 89 ••• [SSM] The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years. they would produce all of Shakespeare’s works. Romans.16 × 107 s ⎟ ⎟ ⎝ ⎠ ( ) ≈ 10484 y T TRussell or T ≈ 10 478 TRussell = 10484 y = 10478 6 10 y . countrymen! Lend me your ears. so we have a grand total of 30 characters to choose from. So let it be with Caesar. Assuming the monkeys type at random. period. And. there are then 30330 different possible arrangements of the character set to form a fragment this long. it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want.) Picture the Problem There are 26 letters and four punctuation marks (space. . The evil that men do lives on after them. I come to bury Caesar. if so. comma. express the probability P that one monkey will write out this passage: Use the approximation 30 ≈ 1000 = 101. This fragment is 330 characters (including spaces) long. Find the time T required for a million monkeys to type this particular passage by accident: Express the ratio of T to Russell’s estimate: P= 1 30330 P= 10 (1.5 )(330 ) 1 = 1 = 10 −495 495 10 T= (330 s )(10495 ) 106 ⎛ ⎞ 1y = 3. it were a grievous fault. Even with this small fragment. The good is oft interred with the bones. (You can even assume that the monkeys are immortal. it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. and exclamation point) used in the English language.30 × 10 491 s ⎜ ⎜ 3. disregarding capitalization. The noble Brutus hath told you that Caesar was ambitious.1862 Chapter 19 Friends. not to praise him. And grievously hath Caesar answered it . . We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare.5 to obtain: Assuming the monkeys can type at a rate of 1 character per second.